Codeforces Round #512 (Div. 2, based on Technocup 2019 Elimination Round 1) C. Vasya and Golden Ticket 【。。。】

任意门：http://codeforces.com/contest/1058/problem/C

C. Vasya and Golden Ticket

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Recently Vasya found a golden ticket — a sequence which consists of nn digits a1a2…ana1a2…an. Vasya considers a ticket to be lucky if it can be divided into two or more non-intersecting segments with equal sums. For example, ticket 350178350178 is lucky since it can be divided into three segments 350350, 1717 and 88: 3+5+0=1+7=83+5+0=1+7=8. Note that each digit of sequence should belong to exactly one segment.

Help Vasya! Tell him if the golden ticket he found is lucky or not.

Input

The first line contains one integer nn (2≤n≤1002≤n≤100) — the number of digits in the ticket.

The second line contains nn digits a1a2…ana1a2…an (0≤ai≤90≤ai≤9) — the golden ticket. Digits are printed without spaces.

Output

If the golden ticket is lucky then print "YES", otherwise print "NO" (both case insensitive).

Examples

input

Copy

5
73452

output

Copy

YES

input

Copy

4
1248

output

Copy

NO

Note

In the first example the ticket can be divided into 77, 3434 and 5252: 7=3+4=5+27=3+4=5+2.

In the second example it is impossible to divide ticket into segments with equal sum.

题意概括：

一串长度为N的数字，判断是否能分成相同和的若干块。

解题思路：

记一道被自己蠢哭的水题，主要是理解题意的问题；

一开始以为分块可以不连续，又是排序又乱七八糟搞一大堆；

结果分块是连续的，活生生的贪心水题。

读题很重要。

AC code：

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 using namespace std;
 const int MAXN = ;
 char str[MAXN];
 int N, sum;
 int main()
 {
     scanf("%d", &N);
     scanf("%s", &str);
     bool flag = true;
     for(int i = ; i < N-; i++){
         sum += str[i]-'';
         int pos = i+;
         flag = true;
         while(pos < N){
             int sum2 = str[pos++]-'';
             while(pos < N && sum2+str[pos]-'' <= sum){
                 sum2+=str[pos++]-'';
             }
             if(sum2 != sum) {flag = false; break;}
         }
         if(flag) {puts("YES"); break;}
     }
     if(flag == false) puts("NO");
     return ;
 }