Problem: You are given two integers n and d. You need to construct a rooted binary tree consisting of n vertices with a root at the vertex 1 and the sum of depths of all vertices equals to d.

A tree is a connected graph without cycles. A rooted tree has a special vertex called the root. A parent of a vertex v is the last different from v vertex on the path from the root to the vertex v. The depth of the vertex v is the length of the path from the root to the vertex v. Children of vertex v are all vertices for which v is the parent. The binary tree is such a tree that no vertex has more than 2 children.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

The only line of each test case contains two integers n and d (2 ≤ n, d ≤ 5000) — the number of vertices in the tree and the required sum of depths of all vertices.

It is guaranteed that the sum of n and the sum of d both does not exceed 5000 (∑ n ≤ 5000, ∑ d ≤ 5000).

Output

For each test case, print the answer.

If it is impossible to construct such a tree, print "NO" (without quotes) in the first line. Otherwise, print "{YES}" in the first line. Then print n-1 integers p_2, p_3, ..., p_n in the second line, where p_i is the parent of the vertex i. Note that the sequence of parents you print should describe some binary tree.

Example

Input


3
5 7
10 19
10 18


Output


YES
1 2 1 3 
YES
1 2 3 3 9 9 2 1 6 
NO

Note

Pictures corresponding to the first and the second test cases of the example:

<image>

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAX = 5e3 + 10;
int T, N, D;
int num[MAX], ans[MAX];
void solve() {
  if (D > (N - 1) * N / 2) {
    printf("NO\n");
    return;
  }
  memset(num, 0, sizeof(num));
  num[0] = 1;
  for (int i = 2; i <= N; i++)
    for (int j = 1; j <= N; j++)
      if (num[j] < 2 * num[j - 1]) {
        num[j]++;
        D -= j;
        break;
      }
  if (D < 0) {
    printf("NO\n");
    return;
  }
  while (D--) {
    for (int i = 1; i <= N; i++)
      if (num[i] < 2 * (num[i - 1] - 1)) {
        num[i]++, num[i - 1]--;
        break;
      }
  }
  printf("YES\n");
  int tot = 1, now = 1;
  for (int i = 1; i <= N; i++) {
    int pre = now + 1;
    for (int j = 0; j < num[i]; j++) ans[++now] = tot + j / 2;
    tot = pre;
  }
  for (int i = 2; i <= N; i++) printf("%d%s", ans[i], i == N ? "\n" : " ");
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d", &N, &D);
    solve();
  }
  return 0;
}