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 = 5009;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
int f[10000];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    memset(f, 0, sizeof(f));
    int n, d, dmin = 0, dmax;
    scanf("%d%d", &n, &d);
    dmax = (n - 1) * n / 2;
    int k = 1, tem = 1;
    while (tem < n) {
      tem += (1 << k);
      dmin += k * (1 << k);
      k++;
    }
    dmin = dmin - (((1 << (k - 1)) - (n - (tem - (1 << (k - 1))))) * (k - 1));
    if (d > dmax || d < dmin)
      printf("NO\n");
    else {
      printf("YES\n");
      if (d == dmax) {
        for (int i = 1; i <= n - 1; i++) printf("%d ", i);
        printf("\n");
        continue;
      }
      k--;
      int down = (n - 1), up = 1;
      for (int i = 1; i <= n - 1; i++) f[i] = 1;
      int flag = 0;
      for (int i = (n - 1); i >= k; i--) {
        int z = i;
        while (1) {
          if (dmax == d) {
            flag = 1;
            break;
          }
          if (z == up) break;
          if (f[up] == (1 << up)) up++;
          dmax--;
          f[z]--;
          f[z - 1]++;
          z--;
        }
        if (flag) break;
      }
      int op;
      for (int i = 1; i <= n - 1; i++) {
        if (f[i] == 0) {
          op = i;
          break;
        }
      }
      op--;
      int p = 1, np = 1, cnt;
      for (int i = 1; i <= op; i++) {
        int t = p;
        cnt = 0;
        for (int j = 1; j <= f[i]; j++) {
          p++;
          cnt++;
          if (cnt >= 3) {
            np++;
            cnt = 1;
          }
          printf("%d ", np);
        }
        np = t + 1;
      }
      printf("\n");
    }
  }
  return 0;
}