題干：

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.

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題目大意：

給你 n 個整數數列 a1,a2,…an?，每個數列的長度為Li。

請你找出兩個編號不同的數列，并從這兩個數列中各恰好刪除一個數，使得這兩個數列的和相等。

解題報告：

? 首先看到讀入這么多數據，所以肯定不能用數組保存下來在離線處理，肯定是邊讀入邊判斷的，這種題通常是讀入的同時，用數據結構記錄一些信息，然后用后面讀入的數去判斷的。這樣就很容易想到，需要記錄的信息就是有哪些數字是可用的，題意都不用轉化，模擬題意就行了：刪去一個數求和。所以先記錄下sum，然后再減去每一個數，用map記錄下是第幾個序列的第幾個位置，方便后來查詢的時候輸出答案。

? 然后為了處理不在同一個序列中，這個問題，剛開始想把序列離散化了，后來想了想不行，，因為還需要記錄當前數的位置，所以干脆直接判斷原來那個數是否也是第i個序列。或者還有一種處理方法

AC代碼：（156ms，改成uset則124ms）

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; typedef pair<int,int> PII;//pair<第i個序列 , 第j個字符> const int MAX = 2e5 + 5; int a[MAX]; map<int,PII> mp; set<int> ss; int main() {int n,len,flag = 0;int ans1,ans2,ans3,ans4;cin>>n;for(int i = 1; i<=n; i++) {scanf("%d",&len);int sum = 0;for(int j = 1; j<=len; j++) scanf("%d",a+j),sum += a[j];if(flag) continue;for(int j = 1; j<=len; j++) {int tmp = sum - a[j];if(ss.count(tmp) && mp[tmp].fi != i) {flag = 1;ans1 = mp[tmp].fi; ans2 = mp[tmp].se;ans3 = i; ans4 = j;}else {ss.insert(tmp);mp[tmp] = pm(i,j);}}} if(flag) {puts("YES");printf("%d %d\n%d %d\n",ans1,ans2,ans3,ans4);}else {puts("NO");}return 0 ;}

另一種處理辦法：（78ms）

for(int i = 1; i<=n; i++) {scanf("%d",&len);int sum = 0;for(int j = 1; j<=len; j++) scanf("%d",a+j),sum += a[j];if(flag) continue;for(int j = 1; j<=len; j++) {int tmp = sum - a[j];if(mp.count(sum - a[j])) {flag = 1;ans1 = mp[tmp].fi; ans2 = mp[tmp].se;ans3 = i; ans4 = j;}}for(int j = 1; j<=len; j++) {int tmp = sum - a[j];mp[tmp] = {i,j};}}

錯誤代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; typedef pair<int,int> PII;//pair<第i個序列 , 第j個字符> const int MAX = 2e5 + 5; int a[MAX]; map<int,PII> mp; set<int> ss; int main() {int n,len,flag = 0;int ans1,ans2,ans3,ans4;cin>>n;for(int i = 1; i<=n; i++) {scanf("%d",&len);int sum = 0;for(int j = 1; j<=len; j++) scanf("%d",a+j),sum += a[j];sort(a+1,a+len+1);int x = unique(a+1,a+len+1) - a - 1;// printf("x = %d\n",x);if(flag) continue;for(int j = 1; j<=x; j++) {int tmp = sum - a[j];if(ss.count(tmp)) {flag = 1;ans1 = mp[tmp].fi;ans2 = mp[tmp].se;ans3 = i;ans4 = j;}else {ss.insert(tmp);mp[tmp] = pm(i,j);}}} if(flag) {puts("YES");printf("%d %d\n%d %d\n",ans1,ans2,ans3,ans4);}else {puts("NO");}return 0 ;}

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