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>
const int MAXN = 3e5 + 10;
const double eps = 1e-8;
using namespace std;
struct edge {
  int t, v;
  edge *next;
} e[MAXN << 1], *h[MAXN], *o = e;
void add(int x, int y, int vul) {
  o->t = y;
  o->v = vul;
  o->next = h[x];
  h[x] = o++;
}
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
int _, n, k;
int num[5005];
int f[MAXN];
vector<int> vec[MAXN];
int main() {
  _ = read();
  while (_--) {
    n = read();
    k = read();
    for (int i = 1; i <= 5000; i++) vec[i].clear(), f[i] = num[i] = 0;
    int x = 1ll * (n - 1) * n / 2;
    int dep = 0;
    int m = n;
    for (int i = 1; i <= m; i++) {
      if (n > (1 << (i - 1)))
        num[i] = (1 << (i - 1)), n -= (1 << (i - 1)),
        dep += 1ll * (i - 1) * num[i];
      else {
        num[i] = n;
        dep += 1ll * (i - 1) * num[i];
        break;
      }
    }
    if (k < dep || k > x)
      puts("NO");
    else {
      for (int i = dep + 1; i <= k; i++) {
        for (int j = 5000; j >= 1; j--) {
          if (num[j] <= 1) continue;
          num[j]--;
          num[j + 1]++;
          break;
        }
      }
      puts("YES");
      vec[1].push_back(1);
      int tot = 1;
      for (int i = 2; i <= 5000; i++) {
        if (!num[i]) break;
        int cnt = 0;
        int l = 0;
        for (int j = tot + 1; j <= tot + num[i]; j++) {
          f[j] = vec[i - 1][l];
          cnt++;
          vec[i].push_back(j);
          if (cnt == 2) cnt = 0, l++;
        }
        tot += num[i];
      }
      for (int i = 2; i <= tot; i++) printf("%d ", f[i]);
      printf("\n");
    }
  }
  return 0;
}