Round #429 (Div.2)

Posted 2020-10-03 浅忆

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One day Kefa found n baloons. For convenience, we denote color of i-th baloon as s_i — lowercase letter of the Latin alphabet. Also Kefa has k friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print ?YES?, if he can, and ?NO?, otherwise. Note, that Kefa‘s friend will not upset, if he doesn‘t get baloons at all.

 

Input

The first line contains two integers n and k (1?≤?n,?k?≤?100) — the number of baloons and friends.

Next line contains string s — colors of baloons.

 

Output

Answer to the task — ?YES? or ?NO? in a single line.

You can choose the case (lower or upper) for each letter arbitrary.

 

Examples

 

Input

4 2
aabb

Output

YES

 

Input

6 3
aacaab

Output

NO

 

Note

In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second.

In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is ?NO?.

题意：Kefa有n个气球，26个小写字母表示颜色，k个朋友，他要把气球分给k个朋友，但是朋友要得到不同的颜色，否则就是NO

 

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <algorithm>
 4 using namespace std;
 5 int a[26];
 6 int n,k,m;
 7 bool cmp(int x,int y){
 8     return x>y;
 9 }
10 int main(){
11     cin>>n>>k;
12     m=k;
13     for(int i=1;i<=n;i++){
14     char x;
15     scanf(" %c",&x);
16     a[x-‘a‘]++;  //标记每种颜色气球的数量
17 18     sort(a,a+26,cmp);  //从大到小排序，例2:a[0]=4
19     a[0]>k?printf("NO\n"):printf("YES\n");  //4>3 NO
20     return 0;
21 }

 

Leha somehow found an array consisting of n integers. Looking at it, he came up with a task. Two players play the game on the array. Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally?

 

Input

First line of input data contains single integer n (1?≤?n?≤?10⁶) — length of the array.

Next line contains n integers a₁,?a₂,?...,?a_n (0?≤?a_i?≤?10⁹).

 

Output

Output answer in single line. "First", if first player wins, and "Second" otherwise (without quotes).

 

Examples

 

Input

4
1 3 2 3

Output

First

Input

2
2 2

 

Output

Second

 

Note

In first sample first player remove whole array in one move and win.

In second sample first player can‘t make a move and lose

n个数和为偶要分情况讨论，为奇first胜，

 1 #include <iostream>
 2 #include <stdio.h>
 3 using namespace std;
 4 int main(){
 5     int n,m,sum=0,t;
 6     scanf("%d",&n);
 7     for(int i=0;i<n;i++){
 8     scanf("%d",&m);
 9     if(m&1) t++;   //奇数
10     sum+=m;
11     }
12     if(sum&1) printf("First\n");   //为奇
13     else {
14     if(t==0) printf("Second\n");
15     else printf("First\n");
16     }
17     return 0;
18 }

 

 or................................................................................or..............................................................................or

 

 1 #include <iostream>
 2 #include <stdio.h>
 3 using namespace std;
 4 int main(){
 5     int n,m,s=0;
 6     scanf("%d",&n);
 7     for(int i=1;i<=n;i++){
 8     scanf("%d",&m);
 9     if(m&1) s=1;
10     }
11     if(s) printf("First\n");
12     else printf("Second\n");
13     return 0;
14 }

 

Leha like all kinds of strange things. Recently he liked the function F(n,?k). Consider all possible k-element subsets of the set [1,?2,?...,?n]. For subset find minimal element in it. F(n,?k) — mathematical expectation of the minimal element among all k-element subsets.

But only function does not interest him. He wants to do interesting things with it. Mom brought him two arrays A and B, each consists of m integers. For all i,?j such that 1?≤?i,?j?≤?m the condition A_i?≥?B_j holds. Help Leha rearrange the numbers in the array A so that the sum is maximally possible, where A‘ is already rearranged array.

Input

First line of input data contains single integer m (1?≤?m?≤?2·10⁵) — length of arrays A and B.

Next line contains m integers a₁,?a₂,?...,?a_m (1?≤?a_i?≤?10⁹) — array A.

Next line contains m integers b₁,?b₂,?...,?b_m (1?≤?b_i?≤?10⁹) — array B.

 

Output

Output m integers a‘₁,?a‘₂,?...,?a‘_m — array A‘ which is permutation of the array A.

 

Examples

 

Input

5
7 3 5 3 4
2 1 3 2 3

Output

4 7 3 5 3

 

Input

7
4 6 5 8 8 2 6
2 1 2 2 1 1 2

Output

2 6 4 5 8 8 6

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <algorithm>
 4 
 5 #define maxn 200001
 6 using namespace std;
 7 int a[maxn],n;
 8 pair<int ,int>b[maxn];
 9 bool cmp(int a,int b){
10     return a>b;
11 }
12 bool cmp2(pair<int,int>a,pair<int,int>b){
13     return a.first<b.first;
14 }
15 bool cmp3(pair<int,int>a,pair<int,int>b){
16         return a.second<b.second;
17 }
18 int main(){
19     cin>>n;
20     for(int i=1;i<=n;i++)
21         cin>>a[i];
22     sort(a+1,a+1+n,cmp);
23     for(int i=1;i<=n;i++){
24     cin>>b[i].first;
25     b[i].second=i;
26     }
27     sort(b+1,b+n+1,cmp2);
28     for(int i=1;i<=n;i++)
29         b[i].first=a[i];
30     sort(b+1,b+1+n,cmp3);
31     for(int i=1;i<=n;i++)
32         cout<<b[i].first<<‘ ‘;
33     return 0;
34 }

 

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