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;
long long n, d, a[100069], dh[100069], pr[100069];
int main() {
  long long t, rr, i, j, k, mx;
  scanf("%lld", &t);
  for (rr = 0; rr < t; rr++) {
    scanf("%lld%lld", &n, &d);
    dh[1] = 0;
    a[0] = 1;
    mx = 0;
    for (i = 2; i <= n; i++) {
      pr[i] = i / 2;
      dh[i] = dh[pr[i]] + 1;
      if (dh[i - 1] < dh[i]) {
        a[dh[i]] = i;
      }
      d -= dh[i];
      mx = dh[i];
    }
    if (d < 0) {
      printf("NO\n");
      continue;
    }
    for (i = n; i - 1; i--) {
      if (a[dh[i]] != i) {
        k = min(mx + 1 - dh[i], d);
        if (k) {
          dh[i] += k;
          pr[i] = a[dh[i] - 1];
          d -= k;
        }
        if (dh[i] > mx) {
          a[dh[i]] = i;
          mx = dh[i];
        }
      }
    }
    if (!d) {
      printf("YES\n");
      for (i = 2; i <= n; i++) {
        printf("%lld%c", pr[i], " \n"[i == n]);
      }
    } else {
      printf("NO\n");
    }
  }
}