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:

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<image>
#include <bits/stdc++.h>
using namespace std;
const int N = 5e3 + 10;
int T, n, d, p[N], dep[N], tree[N][N], cnt[N], sum, Max;
int main() {
  scanf("%d", &T);
  while (T--) {
    memset(p, 0, sizeof(p));
    memset(cnt, 0, sizeof(cnt));
    dep[1] = sum = Max = 0;
    scanf("%d %d", &n, &d);
    tree[0][++cnt[0]] = 1;
    for (int i = 1; i <= n; i++) {
      int LS = i << 1, RS = i << 1 | 1;
      if (LS <= n)
        dep[LS] = dep[i] + 1, tree[dep[LS]][++cnt[dep[LS]]] = LS,
        sum += dep[LS], Max = max(Max, dep[LS]);
      if (RS <= n)
        dep[RS] = dep[i] + 1, tree[dep[RS]][++cnt[dep[RS]]] = RS,
        sum += dep[RS], Max = max(Max, dep[RS]);
    }
    if (sum > d || d > n * (n - 1) / 2) {
      puts("NO");
      continue;
    }
    while (sum < d) {
      for (int i = Max; i >= 0; i--) {
        if (cnt[i] >= 2 && cnt[i + 1] < (cnt[i] - 1) * 2) {
          tree[i + 1][++cnt[i + 1]] = tree[i][cnt[i]], cnt[i]--, sum++;
          Max = max(Max, i + 1);
          break;
        }
      }
    }
    for (int k = 0; k < Max; k++)
      for (int i = 1; i <= cnt[k]; i++) {
        int LS = 2 * (i - 1) + 1, RS = 2 * (i - 1) + 2;
        if (LS <= cnt[k + 1]) p[tree[k + 1][LS]] = tree[k][i];
        if (RS <= cnt[k + 1]) p[tree[k + 1][RS]] = tree[k][i];
      }
    puts("YES");
    for (int i = 2; i <= n; i++) printf("%d ", p[i]);
    puts("");
  }
  return 0;
}
