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 maxn = 5005;
const long long Inf = 1e11;
int p[maxn];
int num[maxn];
int main() {
  int T;
  cin >> T;
  while (T--) {
    memset(p, 0, sizeof(p));
    memset(num, 0, sizeof(num));
    int n, d;
    cin >> n >> d;
    int mi = 0, ma = 0;
    for (int i = 2; i <= n; i++) ma += (i - 1);
    int tn = n - 1, i = 1;
    num[0] = 1;
    while ((1 << i) <= tn) {
      num[i] = (1 << i);
      tn -= (1 << i);
      mi += i * (1 << i);
      i++;
    }
    if (tn) {
      mi += i * tn;
      num[i] = tn;
    }
    int dep = i;
    if (mi <= d && d <= ma) {
      int k = dep;
      int left = d - mi;
      while (left--) {
        for (int i = k; i >= 1; i--) {
          if (num[i] > 1 && 2 * num[i] - 1 > num[i + 1]) {
            num[i]--, num[i + 1]++;
            dep = max(dep, i + 1);
            k = i + 1;
            break;
          }
        }
      }
      int x = 2, pre = 0, son = 0;
      for (int i = 1; i <= dep; i++) {
        pre = x - num[i - 1];
        son = 0;
        for (int j = 1; j <= num[i]; j++) {
          if (son < 2)
            son++, p[x] = pre;
          else
            pre++, son = 1, p[x] = pre;
          x++;
        }
      }
      printf("YES\n");
      for (int i = 2; i <= n; i++) printf("%d ", p[i]);
      printf("\n");
    } else
      printf("NO\n");
  }
}