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 N = 5003;
int T;
int n, m;
int fa[N];
int depth[N];
int sum;
int rep[N];
int main() {
  scanf("%d", &T);
  while (T--) {
    memset(rep, 0, sizeof(rep));
    scanf("%d%d", &n, &m);
    sum = 0;
    for (int i = 2; i <= n; ++i) {
      fa[i] = i / 2;
      depth[i] = depth[fa[i]] + 1;
      sum += depth[i];
    }
    if (m < sum) {
      puts("NO");
      continue;
    }
    for (int i = 1; i <= depth[n]; ++i) rep[i] = (1 << i);
    for (int i = n; i >= 1; --i) {
      if ((i & (i - 1)) == 0) continue;
      while (sum < m) {
        fa[i] = rep[depth[i]];
        ++depth[i];
        ++sum;
        if (!rep[depth[i]]) {
          rep[depth[i]] = i;
          break;
        }
      }
    }
    if (sum < m) {
      puts("NO");
      continue;
    }
    puts("YES");
    for (int i = 2; i <= n; ++i) printf("%d ", fa[i]);
    puts("");
  }
  return 0;
}