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 long long MOD = 1e9 + 7;
const long long MAX = 6000;
long long arr[MAX];
long long rpta[30];
long long ini[MAX];
int main() {
  int t;
  cin >> t;
  long long n, d;
  while (t--) {
    memset(arr, 0, sizeof(arr));
    cin >> n >> d;
    long long ind = 0;
    while (n) {
      arr[ind] = min(n, 1LL << ind);
      n -= arr[ind];
      ind++;
    }
    long long minimo = 0;
    for (long long i = (long long)(0); i < (long long)(ind); i++)
      minimo += i * arr[i];
    if (d < minimo)
      cout << "NO" << endl;
    else {
      long long sobra = d - minimo;
      long long act = ind - 1;
      while (sobra) {
        while (act && arr[act] == 1) act--;
        if (act == 0) break;
        arr[act]--;
        arr[++act]++;
        sobra--;
      }
      if (sobra)
        cout << "NO" << endl;
      else {
        cout << "YES" << endl;
        ini[0] = 1;
        long long sig = 2;
        long long ind = 1;
        bool imp = false;
        bool primera;
        while (arr[ind]) {
          primera = true;
          for (long long i = (long long)(0); i < (long long)(arr[ind]); i++) {
            if (imp)
              cout << " ";
            else
              imp = true;
            if (primera) ini[ind] = sig, primera = false;
            sig++;
            long long padre = ini[ind - 1] + (i / 2);
            cout << padre;
          }
          ind++;
        }
        cout << endl;
      }
    }
  }
}