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 INF = 0x3f3f3f3f, N = 5000 + 5;
int n, m;
int ce[N], lim[N], tot, s;
int fa[N];
bool add() {
  bool f = 0;
  for (int i = tot; i > 0; i--) {
    if ((ce[i] - 1) * 2 >= ce[i + 1] + 1) {
      ce[i]--;
      ce[i + 1]++;
      if (i + 1 > tot) tot = i + 1;
      s++;
      f = 1;
      break;
    }
  }
  return f;
}
void init() {
  s = 0;
  tot = 0;
  memset(ce, 0, sizeof(ce));
  memset(fa, 0, sizeof(fa));
}
void print() {
  for (int i = 1; i <= tot; i++) cout << ce[i] << " ";
  cout << endl;
}
int main() {
  int i, T, j;
  cin >> T;
  lim[0] = 1;
  for (i = 1; i <= 5000; i++) {
    if (i >= 20)
      lim[i] = lim[i - 1];
    else
      lim[i] = lim[i - 1] * 2;
  }
  while (T--) {
    cin >> n >> m;
    init();
    n--;
    ce[0] = 1;
    int left = n;
    for (i = 1; i <= n; i++) {
      tot = i;
      if (left - lim[i] > 0) {
        left -= lim[i];
        ce[i] = lim[i];
        s += lim[i] * i;
      } else {
        ce[i] += left;
        s += left * i;
        break;
      }
    }
    if (s > m) {
      cout << "NO" << endl;
      continue;
    }
    bool f = 1;
    while (s < m) {
      if (!add()) {
        f = 0;
        break;
      }
    }
    if (!f) {
      printf("NO\n");
      continue;
    }
    printf("YES\n");
    int num = 0;
    vector<int> v[2];
    v[0].clear();
    v[1].clear();
    v[1].push_back(1);
    num = 1;
    for (i = 1; i <= tot; i++) {
      v[(i & 1) ^ 1].clear();
      for (j = 1; j <= ce[i]; j++) {
        num++;
        fa[num] = v[i & 1][(j - 1) / 2];
        v[(i & 1) ^ 1].push_back(num);
      }
    }
    for (i = 2; i <= n + 1; i++) cout << fa[i] << " ";
    cout << endl;
  }
}