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:

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<image>
#include <bits/stdc++.h>
using namespace std;
const int N = 5e3 + 5;
const int INF = 0x3f3f3f3f;
int t, n, d, lf[N], rt[N], nok[N], p[N], h[N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> d;
    int lb = 0, aux = 0;
    for (int i = 1; i <= n; i++) {
      lf[i] = rt[i] = nok[i] = p[i] = h[i] = 0;
      lb += aux;
      if (!(i & (i + 1))) aux++;
    }
    for (int i = 1; i + 1 <= n; i++) p[i + 1] = i, h[i + 1] = i, lf[i] = i + 1;
    int ans = (n * (n - 1)) / 2;
    if (ans < d or d < lb)
      cout << "NO\n";
    else {
      int cond = 1;
      while (ans != d) {
        int mn = INF, node = -1;
        for (int i = 1; i <= n; i++)
          if (!lf[i] and !rt[i] and !nok[i] and h[i] < mn) mn = h[i], node = i;
        if (node == -1) {
          cond = 0;
          break;
        }
        int nxt = -1;
        for (int i = 1; i <= n; i++)
          if (h[i] == mn - 2 and (!lf[i] or !rt[i])) {
            nxt = i;
            break;
          }
        if (nxt != -1) {
          if (lf[p[node]] == node)
            lf[p[node]] = 0;
          else
            rt[p[node]] = 0;
          p[node] = nxt;
          if (!lf[nxt])
            lf[nxt] = node;
          else
            rt[nxt] = node;
          h[node] = h[nxt] + 1;
          ans--;
        } else
          nok[node] = 1;
      }
      if (!cond)
        cout << "NO\n";
      else {
        cout << "YES\n";
        for (int i = 2; i <= n; i++) cout << p[i] << " ";
        cout << "\n";
      }
    }
  }
}
