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 maxn = 5005;
const int INF = 0x3f3f3f3f;
int Dep[maxn], sz[maxn];
int main() {
  int t;
  scanf("%d", &t);
  long long cnt = 1;
  for (int i = 0; i < 5000; ++i) {
    sz[i] = cnt;
    if (cnt >= INF)
      continue;
    else
      cnt *= 2;
  }
  while (t--) {
    int n, m, Md = 0;
    scanf("%d%d", &n, &m);
    memset(Dep, 0, sizeof(Dep));
    long long cnt = 0, sum = 0;
    for (int c = 1, d = 0;; c *= 2, d++) {
      Md = max(Md, d);
      if (cnt + c >= n) {
        Dep[d] = n - cnt;
        sum += Dep[d] * d;
        break;
      } else {
        Dep[d] = c;
        cnt += c;
        sum += Dep[d] * d;
      }
    }
    if (sum > m)
      printf("NO\n");
    else {
      bool flag = true;
      for (int i = sum + 1; i <= m; ++i) {
        bool ok = false;
        for (int j = Md; j >= 0; --j) {
          if ((Dep[j] > 1) && ((Dep[j] - 1) * 2 >= Dep[j + 1] + 1) &&
              (Dep[j + 1] + 1 <= sz[j + 1])) {
            Dep[j]--;
            Dep[j + 1]++;
            Md = max(Md, j + 1);
            ok = true;
            break;
          }
        }
        if (!ok) {
          flag = false;
          break;
        }
      }
      if (!flag)
        printf("NO\n");
      else {
        queue<int> fa;
        queue<int> son;
        fa.push(1);
        int idx = 1;
        printf("YES\n");
        for (int i = 1; i <= Md; ++i) {
          for (int j = 0; j < Dep[i]; ++j) {
            if (j && (j % 2) == 0) fa.pop();
            printf("%d ", fa.front());
            son.push(++idx);
          }
          fa = son;
          while (!son.empty()) son.pop();
        }
        printf("\n");
      }
    }
  }
  return 0;
}