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 maxn = 5e3 + 10;
int l[maxn], r[maxn], p[maxn];
int nivel[maxn];
bool mark[maxn];
int main(void) {
  int tc;
  scanf("%d", &tc);
  while (tc--) {
    int n, d;
    scanf("%d %d", &n, &d);
    for (int i = 1; i <= n; i++) mark[i] = 0;
    if (d > n * (n - 1) / 2) {
      printf("NO\n");
      continue;
    }
    int mn = 0;
    int j = 1;
    for (int i = 1; i <= 14; i++) {
      mn += i * min(n - j, (1 << i));
      if (n - j <= (1 << i)) break;
      j += (1 << i);
    }
    if (d < mn) {
      printf("NO\n");
      continue;
    }
    p[1] = 0;
    for (int i = 1; i < n; i++) {
      l[i] = i + 1, r[i] = -1;
      p[i + 1] = i, nivel[i] = i - 1;
    }
    l[n] = r[n] = -1;
    nivel[n] = n - 1;
    int soma = n * (n - 1) / 2;
    while (soma > d) {
      int mn = n + 1;
      int u = -1, v = -1;
      for (int i = 1; i <= n; i++)
        if (l[i] == -1 && r[i] == -1 && !mark[i] && nivel[i] < mn)
          u = i, mn = nivel[i];
      if (u == -1) break;
      for (int i = 1; i <= n; i++)
        if (nivel[i] == nivel[u] - 2 && (l[i] == -1 || r[i] == -1)) v = i;
      if (v == -1) {
        mark[u] = 1;
        continue;
      }
      soma--;
      if (l[p[u]] == u)
        l[p[u]] = -1;
      else
        r[p[u]] = -1;
      nivel[u]--;
      if (l[v] == -1)
        l[v] = u;
      else
        r[v] = u;
      p[u] = v;
    }
    printf("YES\n");
    for (int i = 2; i <= n; i++) printf("%d ", p[i]);
    printf("\n");
  }
}