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;
int n, d, range_min, range_max;
vector<int> node[MAXN];
int fth[MAXN], dep[MAXN], num[MAXN];
signed main() {
  int ts;
  scanf("%d", &ts);
  while (ts--) {
    scanf("%d %d", &n, &d);
    range_min = 0, range_max = 0;
    for (int N = n, i = 0, j = 1;; i++, j *= 2) {
      if (N > j) {
        N -= j;
        range_min += i * j;
      } else {
        range_min += i * N;
        break;
      }
    }
    for (int i = 1; i <= n; i++) range_max += i - 1;
    if (!(range_min <= d && d <= range_max)) {
      printf("NO\n");
      continue;
    }
    printf("YES\n");
    int sum = 0;
    for (int i = 0; i <= n; i++) {
      node[i].clear();
      fth[i] = dep[i] = num[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
      dep[i] = i - 1;
      node[dep[i]].push_back(i);
    }
    for (int i = 2; i <= n; i++) {
      int it = (int)(node[dep[i]].size() + 1) / 2 - 1;
      fth[i] = node[dep[i] - 1][it];
      num[fth[i]]++;
    }
    for (int i = 1; i <= n; i++) sum += dep[i];
    queue<int> q;
    q.push(n);
    while (!q.empty()) {
      if (sum <= d) break;
      int p = q.front();
      q.pop();
      if (num[p] > 0) continue;
      if (node[dep[p]].back() != p) {
        q.push(p);
        continue;
      }
      if (dep[p] <= 1) continue;
      if (node[dep[p] - 1].size() == 2 * node[dep[p] - 2].size()) {
        q.push(p);
        continue;
      }
      node[dep[p]].pop_back();
      dep[p]--;
      node[dep[p]].push_back(p);
      sum--;
      num[fth[p]]--;
      q.push(fth[p]);
      int it = (int)(node[dep[p]].size() + 1) / 2 - 1;
      fth[p] = node[dep[p] - 1][it];
      num[fth[p]]++;
      q.push(p);
    }
    for (int i = 2; i <= n; i++) printf("%d ", fth[i]);
    printf("\n");
  }
  return 0;
}