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;
int t, n, d, s, mx, a[5001], b[5001], dp[5001], c[5001][5001];
inline long long read() {
  long long sum = 0, x = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') x = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    sum = sum * 10 + ch - '0';
    ch = getchar();
  }
  return sum * x;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
  return;
}
int main() {
  t = read();
  while (t--) {
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));
    dp[1] = s = mx = 0;
    n = read();
    d = read();
    c[0][++b[0]] = 1;
    for (register int i = 1; i <= n; ++i) {
      int l = i << 1, r = i << 1 | 1;
      if (l <= n) {
        dp[l] = dp[i] + 1;
        c[dp[l]][++b[dp[l]]] = l;
        s += dp[l];
        mx = max(mx, dp[l]);
      }
      if (r <= n) {
        dp[r] = dp[i] + 1;
        c[dp[r]][++b[dp[r]]] = r;
        s += dp[r];
        mx = max(mx, dp[r]);
      }
    }
    if (s > d || d > n * (n - 1) / 2) {
      puts("NO");
      continue;
    }
    while (s < d)
      for (register int i = mx; i >= 0; --i)
        if (b[i] >= 2 && b[i + 1] < (b[i] - 1) * 2) {
          c[i + 1][++b[i + 1]] = c[i][b[i]];
          --b[i];
          ++s;
          mx = max(mx, i + 1);
          break;
        }
    for (register int k = 0; k < mx; ++k)
      for (register int i = 1; i <= b[k]; ++i) {
        int l = 2 * (i - 1) + 1, r = 2 * (i - 1) + 2;
        if (l <= b[k + 1]) a[c[k + 1][l]] = c[k][i];
        if (r <= b[k + 1]) a[c[k + 1][r]] = c[k][i];
      }
    puts("YES");
    for (register int i = 2; i <= n; ++i) {
      write(a[i]);
      putchar(' ');
    }
    putchar('\n');
  }
  return 0;
}