codeforces 754A（思维题）

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 arrayA 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 arrayA.

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 arrayA.

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 integerk — the number of new arrays. In each of the nextk lines print two integers l_i and r_i which denote the subarrayA[l_i...r_i] of the initial array A being the i-th new array. Integersl_i,r_i should satisfy the following conditions:

• l₁ = 1
• r_k = n
• r_i + 1 = l_i + 1 for each1 ≤ i < k.

If there are multiple answers, print any of them.

Example

Input

31 2 -3

Output

YES21 23 3

Input

89 -12 3 4 -4 -10 7 3

Output

YES21 23 8

Input

10

Output

NO

Input

41 2 3 -5

Output

YES41 12 23 34 4

/*题意让你将一个数组任意拆分为多个数组，按照原来的元素的顺序，满足的条件是每组数组的和不能为0，这些数组元素必须是原来的顺序一个接一个，可能会有多种解人以输出一个即可*/

/*思路：最简单的方法就是将原来的数组拆分为的数组只含有一个元素，保证每个元素不为零，需要特判，可能原来数组有多个0，都为0，前半部分为0,后半部分为0，中间部分为0的情况。详情见代码*/

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int num[200];int main(){    int n;    while(scanf("%d", &n) != EOF){        memset(num, 0, sizeof(num));        int cnt = 0;        for(int i = 1; i <= n; i++){            scanf("%d", &num[i]);            if(num[i] == 0) ++cnt;        }        int flag = 0;        int f0 = 0;        if(cnt == n){    //按照这种思路输出为no的情况只有所有元素为0（cnt统计的是0的个数）            printf("NO\n");            continue;        }        else{            printf("YES\n");            int t = n - cnt;            printf("%d\n", t);            for(int i = 1; i <= n; i++){                if(t == 1) {    //这里表示当前只有一个数字不为零其他都为0那么只用输出一组即可                    printf("%d %d\n", i, n);                    break;                }                else {                    if(num[i] == 0){                        if(!f0){    //若当前为0，直到下一个非0时才能停止那么需要标记                            printf("%d ", i);                            f0 = 1;                        }                        flag = 1;                        continue;                    }                    if(flag){                        printf("%d\n", i);                        flag = 0;                        f0 = 0;                        t--;    //这里的原因是为了解决若有多个0                        continue;                    }                    else{                        printf("%d %d\n", i, i);                        t--;                    }                }            }        }    }    return 0;}