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 = 5e3 + 2;
int d[N], p[N], c[N];
void solve(int n, int x) {
  int sum = 0;
  for (int i = 0; i < n; i++) {
    sum += i;
    d[i + 1] = i;
    p[i + 1] = i;
    c[i] = 1;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= i - 2; j++) {
      int depth_change = (d[i] - (d[j] + 1)) * (n - i + 1);
      if (c[j] < 2 && d[j] + 1 < d[i] && sum - depth_change >= x) {
        c[j]++;
        c[p[i]]--;
        p[i] = j;
        sum -= depth_change;
        for (int k = i; k <= n; k++) {
          d[k] = (d[j] + 1) + k - i;
        }
        break;
      }
    }
  }
  if (sum == x) {
    puts("YES");
    for (int i = 2; i <= n; i++) {
      printf("%d%c", p[i], " \n"[i == n]);
    }
  } else {
    puts("NO");
  }
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, x;
    scanf("%d%d", &n, &x);
    solve(n, x);
  }
  return 0;
}