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 = (long long)1e9 + 7;
const long double PI = 3.141592653589793238462643383279502884197;
priority_queue<int, vector<int>, greater<int>> pq;
vector<int> v[5001];
int n, d;
int num[5001];
int cap[5001];
int pp[5001];
int main() {
  int tc;
  scanf("%d", &tc);
  while (tc--) {
    fill(cap, cap + 5001, 2);
    memset(num, 0, sizeof(num));
    scanf("%d %d", &n, &d);
    int mav = n * (n - 1) / 2;
    for (int i = 0; i < n; i++) v[i].clear();
    if (d > mav) {
      printf("NO\n");
      continue;
    }
    int miv = 0, cnt = 0, t = 1, cmd = 0;
    for (int i = 0; i < n; i++) {
      cnt++;
      num[cmd]++;
      miv += cmd;
      if (t == cnt) {
        t *= 2;
        cnt = 0;
        cmd++;
      }
    }
    if (miv > d) {
      printf("NO\n");
      continue;
    }
    for (int i = d - miv - 1; i >= 0; i--) {
      for (int j = 1; j < n; j++) {
        if ((num[j] - 1) * 2 >= (num[j + 1] + 1)) {
          num[j]--;
          num[j + 1]++;
          break;
        }
      }
    }
    v[0].push_back(1);
    cnt = 1;
    printf("YES\n");
    for (int i = 1; i < n; i++) {
      while (num[i]--) {
        cnt++;
        v[i].push_back(cnt);
        for (int j : v[i - 1]) {
          if (cap[j]) {
            cap[j]--;
            printf("%d ", j);
            break;
          }
        }
      }
    }
    printf("\n");
  }
}