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>
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
const int MAXN = 5e3 + 5;
int l[MAXN], r[MAXN], fa[MAXN];
inline void solve(int now, int N, int D) {
  if (N == 1) return;
  if (l[N - 1] <= D - (N - 1) && r[N - 1] >= D - (N - 1)) {
    fa[now + 1] = now;
    solve(now + 1, N - 1, D - (N - 1));
    return;
  }
  for (int i = 1; i < N - 1; i++)
    for (int j = 0; j <= D - (N - 1); j++)
      if (l[i] <= j && j <= r[i] && l[N - 1 - i] <= D - (N - 1) - j &&
          D - (N - 1) - j <= r[N - 1 - i]) {
        fa[now + 1] = now;
        solve(now + 1, i, j);
        fa[now + i + 1] = now;
        solve(now + i + 1, N - 1 - i, D - (N - 1) - j);
        return;
      }
}
inline void Init() {
  for (int i = 1; i < MAXN; i++) {
    l[i] = l[i - 1] + (int)log2(i);
    r[i] = i * (i - 1) / 2;
  }
}
int main() {
  Init();
  int T = read();
  while (T--) {
    int N = read(), D = read();
    if (D < l[N] || D > r[N]) {
      puts("NO");
      continue;
    }
    solve(1, N, D);
    puts("YES");
    for (int i = 2; i <= N; i++) printf("%d%c", fa[i], " \n"[i == N]);
  }
  return 0;
}
