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 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, ..., an ( - 103 ≤ ai ≤ 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 li and ri which denote the subarray A[li... ri] of the initial array A being the i-th new array. Integers li, rishould satisfy the following conditions:

• l1 = 1
• rk = n
• ri + 1 = li + 1 for each 1 ≤ 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

解题说明：此题是一道水题，只需要保证数组中每一段不为0即可，首先判断数组和是否为0，如果不为0直接满足条件，否则判断是否每个数都为0，不是至少可以分为2段。

#include<cstdio>#include<algorithm>#include<cstring>#include<cstdlib>#include<iostream>using namespace std;int main(){int n, i, a[100], sum = 0;scanf ("%d", &n);for (i = 0; i < n; i++) {scanf ("%d", &a[i]);sum += a[i];}if (sum != 0){printf ("YES\n1\n1 %d", n);}else {for (i = 0; i < n && a[i] == 0; i++);if (i == n){printf ("NO\n");}else{printf ("YES\n2\n%d %d\n%d %d\n", 1, i + 1, i + 2, n);}}return 0;}