Equal Sums (map的基本应用) 多学骚操作

C. Equal Sums

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given $\mathrm{kk sequences\; of\; integers.\; The\; length\; of\; the ii-th\; sequence\; equals\; to nini.}$

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

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 $\mathrm{00)\; sequence\; is 00.}$

Input

The first line contains an integer $\mathrm{kk (2\le k\le 2\cdot 1052\le k\le 2\cdot 105)\; —\; the\; number\; of\; sequences.}$

Then $\mathrm{kk pairs\; of\; lines\; follow,\; each\; pair\; containing\; a\; sequence.}$

The first line in the $\mathrm{ii-th\; pair\; contains\; one\; integer nini (1\le ni<2\cdot 1051\le ni<2\cdot 105)\; —\; the\; length\; of\; the ii-th\; sequence.\; The\; second\; line\; of\; the ii-th\; pair\; contains\; a\; sequence\; of nini integers ai,1,ai,2,\dots ,ai,niai,1,ai,2,\dots ,ai,ni.}$

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

The sum of lengths of all given sequences don't exceed $\mathrm{2\cdot 1052\cdot 105,\; i.e. n1+n2+\cdots +nk\le 2\cdot 105n1+n2+\cdots +nk\le 2\cdot 105.}$

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 $\mathrm{ii, xx (1\le i\le k,1\le x\le ni1\le i\le k,1\le x\le ni),\; in\; the\; third\; line\; —\; two\; integers jj, yy (1\le j\le k,1\le y\le nj1\le j\le k,1\le y\le nj).\; It\; means\; that\; the\; sum\; of\; the\; elements\; of\; the ii-th\; sequence\; without\; the\; element\; with\; index xx equals\; to\; the\; sum\; of\; the\; elements\; of\; the jj-th\; sequence\; without\; the\; element\; with\; index yy.}$

Two chosen sequences must be distinct, i.e. $\mathrm{i\ne ji\ne j.\; You\; can\; print\; them\; in\; any\; order.}$

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

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
struct node {
    int x, y;
    node(int x, int y): x(x), y(y) {}
    node () {}
    bool operator < (node const &a ) const  {
        return a.x < x;
    }
};
map<int,node>mp;
int t, n, a[maxn];
int main() {
    scanf("%d", &t);
    int  flag = 0;
    for (int k = 1 ; k <= t ; k++) {
        scanf("%d", &n);
        int sum = 0;
        for (int i = 0 ; i < n ; i++ ) {
            scanf("%d", &a[i]);
            sum += a[i];
        }
        if (flag) continue;
        for (int i = 0 ; i < n ; i++) {
            int temp = sum - a[i];
            if (mp.find(temp) != mp.end()) {
                if (mp[temp].x != k) {
                    printf("YES
");
                    printf("%d %d
", k, i+1);
                    printf("%d %d
", mp[temp].x, mp[temp].y+1);
                    flag = 1;
                }
            }
            if (flag) break;
            mp[temp] = node(k, i);
        }
    }
    if (!flag) printf("NO
");
    return 0;
}