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 = 5000 + 10;
int n, x, fa[N], ch[N][2], d[N], dt[N];
bool use[N];
int main() {
  int t;
  scanf("%d", &t);
  for (int T = 1; T <= t; T++) {
    scanf("%d%d", &n, &x);
    dt[0] = -1;
    long long cur = n * (n - 1ll) / 2, dl = 0;
    for (int i = 1; i <= n; i++) {
      fa[i] = d[i] = i - 1;
      dt[i] = dt[i >> 1] + 1;
      dl += dt[i];
      ch[i][0] = 0;
      ch[i][1] = i == n ? 0 : i + 1;
    }
    fill(use + 1, use + n + 1, 0);
    bool ok = 1;
    if (x > cur || x < dl) {
      ok = 0;
    } else {
      while (cur > x) {
        int u = -1;
        for (int i = 2; i <= n; i++)
          if (!use[i] && !ch[i][0] && !ch[i][1]) {
            u = u == -1 ? i : (d[i] < d[u] ? i : u);
          }
        if (u == -1) {
          ok = 0;
          break;
        }
        bool tag = 0;
        for (int i = 1; i <= n; i++)
          if (d[i] == d[u] - 2 && (!ch[i][0] || !ch[i][1])) {
            ch[fa[u]][ch[fa[u]][1] == u] = 0;
            fa[u] = i;
            ch[i][ch[i][1] == 0] = u;
            d[u]--;
            cur--;
            tag = 1;
            break;
          }
        if (!tag) use[u] = 1;
      }
    }
    if (!ok)
      puts("NO");
    else {
      puts("YES");
      for (int i = 2; i <= n; i++) printf("%d%c", fa[i], " \n"[i == n]);
    }
  }
  return 0;
}
