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 = 5000 + 5;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, d;
int lef[N], rig[N];
inline void prework() {
  int Max = 5000;
  lef[0] = rig[0] = 0;
  lef[1] = rig[1] = 1;
  for (int i = 2; i <= Max; i++) rig[i] = i * (i + 1) / 2;
  for (int i = 2; i <= Max; i++) {
    int deep = 1;
    while ((1LL << deep) - 1 < i) deep++;
    lef[i] = lef[i - 1] + deep;
  }
}
int fa[N];
inline void Work(int l, int r, int dis) {
  if (r - l <= 0) return;
  int cnt = r - l + 1;
  dis -= cnt;
  for (int i = 1; i <= cnt - 1; i++) {
    int l1 = lef[i], r1 = rig[i];
    int l2 = lef[cnt - 1 - i], r2 = rig[cnt - 1 - i];
    if (dis < l1 + l2 || dis > r1 + r2) continue;
    for (int j = l1; j <= r1; j++) {
      if (dis - j >= l2 && dis - j <= r2) {
        fa[l + 1] = l;
        if (l + i + 1 <= r) fa[l + i + 1] = l;
        Work(l + 1, l + i, j);
        Work(l + i + 1, r, dis - j);
        return;
      }
    }
  }
}
int main() {
  prework();
  int T = read();
  while (T--) {
    n = read();
    d = read();
    d += n;
    if (d < lef[n] || d > rig[n])
      printf("NO\n");
    else {
      printf("YES\n");
      Work(1, n, d);
      for (int i = 2; i <= n; i++) printf("%d ", fa[i]);
      printf("\n");
    }
  }
  return 0;
}