Codeforces 754A Lesha and array splitting(简单贪心)

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, ..., 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

YES 
4 
1 1 
2 2 
3 3 
4 4
题目链接：http://codeforces.com/contest/754/problem/A
分析：求前缀和; 看看pre[n]等于多少; pre[n]!=0; 则直接整个数组全部输出; 如果pre[n]==0 则在前面找一个i pre[i]!=0 如果找到了 则 

输出a[1..i]和a[i+1..n]; 可以看成是pre[0]=0,pre[i]!=0,pre[n]=0 则可知这两段都是符合要求的不为0; 但是如果没有找到pre[i]!=0 

那么就意味着pre[1..n-1]都为0;则数字全为0;则不可能了; 贪心吧。
下面给出AC代码：

#include<iostream>
using namespace std;
int n,a[101],i,s,b;
int main()
{
    cin>>n;
    for(i=1;i<=n;i++)
    {
        cin>>a[i];
        s+=a[i];
        if(a[i])
           b=i;
    }
    if(b==0)
       cout<<"NO\n";
    else if(s)
       cout<<"YES\n1\n"<<1<<" "<<n<<"\n";
    else
       cout<<"YES\n2\n"<<1<<" "<<b-1<<"\n"<<b<<" "<<n<<"\n";
    return 0;
}