CodeForces - 988C Equal Sums

You are given kk sequences of integers. The length of the ii-th sequence equals to nini.

You have to choose exactly two sequences ii and jj (i≠ji≠j) such that you can remove exactly one element in each of them in such a way that the sum of the changed sequence ii (its length will be equal to ni−1ni−1) equals to the sum of the changed sequence jj (its length will be equal to nj−1nj−1).

Note that it's required to remove exactly one element in each of the two chosen sequences.

Assume that the sum of the empty (of the length equals 00) sequence is 00.

Input

The first line contains an integer kk (2≤k≤2⋅1052≤k≤2⋅105) — the number of sequences.

Then kk pairs of lines follow, each pair containing a sequence.

The first line in the ii-th pair contains one integer nini (1≤ni<2⋅1051≤ni<2⋅105) — the length of the ii-th sequence. The second line of the ii-th pair contains a sequence of nini integers ai,1,ai,2,…,ai,niai,1,ai,2,…,ai,ni.

The elements of sequences are integer numbers from −104−104 to 104104.

The sum of lengths of all given sequences don't exceed 2⋅1052⋅105, i.e. n1+n2+⋯+nk≤2⋅105n1+n2+⋯+nk≤2⋅105.

Output

If it is impossible to choose two sequences such that they satisfy given conditions, print "NO" (without quotes). Otherwise in the first line print "YES" (without quotes), in the second line — two integers ii, xx (1≤i≤k,1≤x≤ni1≤i≤k,1≤x≤ni), in the third line — two integers jj, yy (1≤j≤k,1≤y≤nj1≤j≤k,1≤y≤nj). It means that the sum of the elements of the ii-th sequence without the element with index xx equals to the sum of the elements of the jj-th sequence without the element with index yy.

Two chosen sequences must be distinct, i.e. i≠ji≠j. You can print them in any order.

If there are multiple possible answers, print any of them.

Examples

Input

2
5
2 3 1 3 2
6
1 1 2 2 2 1

Output

YES
2 6
1 2

Input

3
1
5
5
1 1 1 1 1
2
2 3

Output

NO

Input

4
6
2 2 2 2 2 2
5
2 2 2 2 2
3
2 2 2
5
2 2 2 2 2

Output

YES
2 2
4 1

Note

In the first example there are two sequences [2,3,1,3,2][2,3,1,3,2] and [1,1,2,2,2,1][1,1,2,2,2,1]. You can remove the second element from the first sequence to get [2,1,3,2][2,1,3,2] and you can remove the sixth element from the second sequence to get [1,1,2,2,2][1,1,2,2,2]. The sums of the both resulting sequences equal to 88, i.e. the sums are equal.

给你k个序列，找两个序列，使得各去掉一个元素后和相等。

一开始感觉很复杂……因为不管怎么做时间复杂度都感觉不太理想。最后还是直接暴力 用和做键，数列和去掉的数位置作为值封装map，有键相等说明符合要求输出对应就行，最后耗时大概只有限时的一半，奇了怪了。

AC代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <map>
using namespace std;
const int maxn=2e5+10;
int a[maxn];
typedef pair<int,int>P;
int main()
{
    int t,n,sum,flag=0;
    cin>>t;
    map<int,P> m;
    m.clear();
    for(int i=0;i<t;i++)
    {
        cin>>n;
        sum=0;
        for(int j=0;j<n;j++)
        {
            cin>>a[j];
            sum+=a[j];
        }
        for(int j=0;j<n;j++)
        {
            int k=sum-a[j];
            if(m.count(k)&&!flag&&m[k].first!=i+1)
            {
                cout<<"YES\n";
                cout<<i+1<<' '<<j+1<<endl;
                cout<<m[k].first<<' '<<m[k].second<<endl;
                flag=1;
            }
            P p;
            p.first=i+1;
            p.second=j+1;
            m[k]=p;
        }
    }
    if(!flag)
        cout<<"NO\n";
    return 0;
}