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 = 5005 + 5;
int t, n, m, l, r, sum[N], dep[N];
bool solve() {
  fill(dep + 1, dep + n + 1, 1);
  m = n * (n - 1) / 2 - m;
  for (l = 2, r = n; l < r; --r) {
    while (l < r && (m - r + l < 0 || dep[l] == dep[l - 1] * 2)) l++;
    if (l >= r) break;
    m -= r, dep[r]--;
    m += l, dep[l]++;
  }
  return m == 0;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &m);
    if (!solve())
      printf("NO\n");
    else {
      printf("YES\n");
      for (int i = 1; i <= r; ++i) sum[i] = sum[i - 1] + dep[i];
      for (int i = 2; i <= r; ++i) {
        for (int j = 0; j < dep[i]; ++j) printf("%d ", sum[i - 2] + 1 + j / 2);
      }
      printf("\n");
    }
  }
  return 0;
}