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 N = 5e3;
vector<int> lvl[N + 100];
int ext[N + 100], par[N + 100];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    for (int i = 0; i <= N; i++) {
      lvl[i].clear();
      ext[i] = 1;
    }
    int n, d;
    scanf("%d%d", &n, &d);
    int mn = 0, mx = 0;
    for (int i = 1; i <= n; i++) {
      mx += i - 1;
      int cnt = 0;
      int cur = i;
      while (cur) {
        cur /= 2;
        cnt++;
      }
      mn += cnt - 1;
    }
    if (d < mn || d > mx) {
      printf("NO\n");
    } else {
      for (int i = 1; i <= n; i++) {
        lvl[i - 1].push_back(i);
      }
      printf("YES\n");
      int lst = 1;
      for (int i = n; mx > d; i--) {
        lvl[i - 1].pop_back();
        int cur = i - 1;
        int minus = min(cur - lst, mx - d);
        lvl[cur - minus].push_back(i);
        ext[cur - minus]--;
        ext[cur - minus + 1] += 2;
        if (ext[lst] == 0) {
          lst++;
        }
        mx -= minus;
      }
      for (int i = 0;; i++) {
        if (lvl[i].empty()) {
          break;
        }
        for (int sz = lvl[i].size(), j = 0; j < sz; j++) {
          int check = lvl[i + 1].size();
          if (j * 2 < check) {
            par[lvl[i + 1][j * 2]] = lvl[i][j];
          }
          if (j * 2 + 1 < check) {
            par[lvl[i + 1][j * 2 + 1]] = lvl[i][j];
          }
        }
      }
      for (int i = 2; i <= n; i++) {
        printf("%d ", par[i]);
      }
      printf("\n");
    }
  }
}