#include <bits/stdc++.h>
using namespace std;
int main() {
  int t, n, s;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> n >> s;
    long long int maxm = -1, sum = 0, ans = 0, x, idx;
    bool flag = true;
    bool flag1 = true;
    for (int j = 0; j < n; j++) {
      cin >> x;
      sum += x;
      if (x > maxm) {
        maxm = x;
        idx = j + 1;
      }
      if (sum > s) {
        if (flag) {
          sum -= maxm;
          flag = false;
          ans = idx;
        } else {
        }
      }
    }
    if (flag) {
      cout << 0 << endl;
    } else {
      cout << ans << endl;
    }
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

New Year is coming! Vasya has prepared a New Year's verse and wants to recite it in front of Santa Claus.

Vasya's verse contains n parts. It takes a_i seconds to recite the i-th part. Vasya can't change the order of parts in the verse: firstly he recites the part which takes a_1 seconds, secondly — the part which takes a_2 seconds, and so on. After reciting the verse, Vasya will get the number of presents equal to the number of parts he fully recited.

Vasya can skip at most one part of the verse while reciting it (if he skips more than one part, then Santa will definitely notice it).

Santa will listen to Vasya's verse for no more than s seconds. For example, if s = 10, a = [100, 9, 1, 1], and Vasya skips the first part of verse, then he gets two presents.

Note that it is possible to recite the whole verse (if there is enough time). 

Determine which part Vasya needs to skip to obtain the maximum possible number of gifts. If Vasya shouldn't skip anything, print 0. If there are multiple answers, print any of them.

You have to process t test cases.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.

The first line of each test case contains two integers n and s (1 ≤ n ≤ 10^5, 1 ≤ s ≤ 10^9) — the number of parts in the verse and the maximum number of seconds Santa will listen to Vasya, respectively.

The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the time it takes to recite each part of the verse.

It is guaranteed that the sum of n over all test cases does not exceed 10^5.

Output

For each test case print one integer — the number of the part that Vasya needs to skip to obtain the maximum number of gifts. If Vasya shouldn't skip any parts, print 0.

Example

Input


3
7 11
2 9 1 3 18 1 4
4 35
11 9 10 7
1 8
5


Output


2
1
0

Note

In the first test case if Vasya skips the second part then he gets three gifts.

In the second test case no matter what part of the verse Vasya skips.

In the third test case Vasya can recite the whole verse.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int t, n, s;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> n >> s;
    long long int maxm = -1, sum = 0, ans = 0, x, idx;
    bool flag = true;
    bool flag1 = true;
    for (int j = 0; j < n; j++) {
      cin >> x;
      sum += x;
      if (x > maxm) {
        maxm = x;
        idx = j + 1;
      }
      if (sum > s) {
        if (flag) {
          sum -= maxm;
          flag = false;
          ans = idx;
        } else {
        }
      }
    }
    if (flag) {
      cout << 0 << endl;
    } else {
      cout << ans << endl;
    }
  }
  return 0;
}

```