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>
#include <bits/stdc++.h>
using namespace std;
long long par[5010], f[5010];
vector<long long> lev[5010];
int main() {
  long long t, i, j, k, n, d, sum;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld %lld", &n, &d);
    sum = n * (n - 1);
    sum /= 2;
    long long lo = 0, x = n;
    for (i = 0; i < n; i++) {
      if ((1 << i) >= x) {
        lo += i * x;
        break;
      }
      lo += (1 << i) * i;
      x -= (1 << i);
    }
    if (d > sum || d < lo) {
      printf("NO\n");
      continue;
    }
    for (i = 2; i <= n; i++) par[i] = i - 1;
    for (i = 1; i <= n; i++) {
      lev[i].clear();
      lev[i].push_back(i);
      f[i] = 0;
    }
    for (i = n; i > 1; i--) {
      if (sum == d) break;
      for (j = i; j >= 2; j--, sum--) {
        if (lev[j - 1].size() <= f[j - 1]) {
          j++;
          sum++;
          break;
        }
        if (sum == d) {
          break;
        }
      }
      if (j == 1) j = 2, sum++;
      par[i] = lev[j - 1][f[j - 1]];
      f[j - 1]++;
      lev[j].push_back(i);
      lev[j].push_back(i);
    }
    printf("YES\n");
    for (i = 2; i <= n; i++) printf("%lld ", par[i]);
    printf("\n");
  }
  return 0;
}
