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

A. Lesha and array splitting

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

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]);        }    }}