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;
int d[5002], rem[5002], D[5002], par[5002];
vector<int> v[5002];
vector<int> node[5002];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, d;
    scanf("%d%d", &n, &d);
    for (int i = 0; i <= n; i++)
      v[i].clear(), D[i] = 0, rem[i] = 0, node[i].clear();
    long long sum = 0;
    for (int i = 0; i <= n - 1; i++) {
      D[i + 1] = i;
      v[i].emplace_back(i + 1);
      rem[i] = 1;
      sum += i;
    }
    rem[n - 1]++;
    while (sum > d) {
      int lv = 0;
      for (int i = n - 1; i >= 0; i--) {
        if (v[i].size() > 0) {
          lv = i;
          break;
        }
      }
      int u = v[lv].back();
      v[lv].pop_back();
      rem[lv] -= 2;
      bool ok = 0;
      for (int i = 0; i <= n - 1; i++) {
        if (rem[i] > 0 && sum - lv + i + 1 >= d && sum - lv + i + 1 < sum) {
          ok = 1;
          rem[i]--;
          rem[i + 1] += 2;
          D[u] = i + 1;
          sum = sum - lv + i + 1;
          break;
        }
      }
      if (!ok) break;
    }
    if (sum != d) {
      printf("NO\n");
    } else {
      for (int i = 1; i <= n; i++) node[D[i]].emplace_back(i);
      for (int i = 0; i <= n - 1; i++) {
        int l = 0;
        for (int j = 0; j < node[i].size(); j++) {
          int u = node[i][j];
          for (int j = 0; j < 2; j++) {
            if (l >= node[i + 1].size()) break;
            int v = node[i + 1][l];
            par[v] = u;
            l++;
          }
        }
      }
      printf("YES\n");
      for (int i = 2; i <= n; i++) printf("%d ", par[i]);
      printf("\n");
    }
  }
}