Codeforces Round #390(Div. 2)A. Lesha and array splitting【思维】

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本文介绍了一个有趣的数组分割问题，目标是将一个整数数组分割成多个子数组，确保每个子数组的元素之和不为零。文章提供了输入输出样例及解析，并分享了一种有效的解决方案。

One spring day on his way to university Lesha found an array A. Lesha likes to split arrays into several parts. This time Lesha decided to split the array A into several, possibly one, new arrays so that the sum of elements in each of the new arrays is not zero. One more condition is that if we place the new arrays one after another they will form the old array A.

Lesha is tired now so he asked you to split the array. Help Lesha!

Input

The first line contains single integer n (1 ≤ n ≤ 100) — the number of elements in the array A.

The next line contains n integers a₁, a₂, ..., a_n ( - 10³ ≤ a_i ≤ 10³) — the elements of the array A.

Output

If it is not possible to split the array A and satisfy all the constraints, print single line containing "NO" (without quotes).

Otherwise in the first line print "YES" (without quotes). In the next line print single integer k — the number of new arrays. In each of the next k lines print two integers l_i and r_i which denote the subarray A[l_i... r_i] of the initial array A being the i-th new array. Integers l_i, r_i should satisfy the following conditions:

l₁ = 1
r_k = n
r_i + 1 = l_i + 1 for each 1 ≤ i < k.

If there are multiple answers, print any of them.

Examples

Input

3
1 2 -3

Output

YES
2
1 2
3 3

Input

8
9 -12 3 4 -4 -10 7 3

Output

YES
2
1 2
3 8

Input

1
0

Output

NO

Input

4
1 2 3 -5

Output

题目大意：

给你一个数组A，让你将其分成若干个子数组，使得每个数组中的数字和都不是0.

思路：

将a【i】！=0的部分，设定为一个子数组，对应将a【i】==0的部分，归到其他数组中即可。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int a[150];
int l[150];
int r[150];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int flag=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(a[i]!=0)flag=1;
        }
        if(flag==0)
        {
            printf("NO\n");
            continue;
        }
        printf("YES\n");
        int ans=0;
        int cont=0;
        for(int i=1;i<=n;i++)
        {
            if(a[i]!=0)
            {
                cont++;
                l[cont]=i;
                r[cont]=i;
            }
        }
        r[0]=0;
        l[cont+1]=n+1;
        for(int i=1;i<=cont;i++)
        {
            l[i]=r[i-1]+1;
            r[i]=l[i+1]-1;
        }
        printf("%d\n",cont);
        for(int i=1;i<=cont;i++)
        {
            printf("%d %d\n",l[i],r[i]);
        }
    }
}