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;
vector<int> g[maxn];
int num[maxn];
int mn, mx;
int n, d;
int ans[maxn], fa[maxn], id[maxn];
bool run() {
  cin >> n >> d;
  for (int i = 1; i <= n; ++i) g[i].clear();
  int rem = n, dep = 0;
  mn = 0;
  for (int i = 0; i <= n; ++i) {
    dep = i;
    if (rem <= (1 << i)) {
      num[i] = rem;
      mn += rem * (i);
      break;
    }
    rem -= (1 << i);
    mn += (1 << i) * (i);
  }
  mx = n * (n - 1) / 2;
  if (d < mn) return false;
  if (d > mx) return false;
  for (int i = 1; i <= n; ++i) ans[i] = 1;
  int down = n, up = 2;
  while (mx > d) {
    ans[down]--;
    if (mx - d > down - up + 1) {
      ans[up]++;
      mx -= down - up;
      if (ans[up] == (1 << (up - 1))) ++up;
    } else {
      for (int j = down + d - mx; j <= n; ++j) {
        if (ans[j] != (1 << (j - 1))) {
          ++ans[j];
          mx -= down - j;
          break;
        }
      }
    }
    --down;
  }
  return true;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    if (run()) {
      printf("YES\n");
      int cnt = 1;
      for (int i = 1; i <= n; ++i) {
        if (!ans[i]) break;
        for (int j = 0; j < ans[i]; ++j) g[i].push_back(cnt++);
      }
      for (int i = 2; i <= n; ++i) {
        if (!ans[i]) break;
        for (int j = 0; j < ans[i]; ++j) {
          fa[g[i][j]] = g[i - 1][j >> 1];
        }
      }
      for (int i = 2; i <= n; ++i) printf("%d%c", fa[i], i == n ? '\n' : ' ');
    } else
      printf("NO\n");
  }
  return 0;
}