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 n, d;
int get_mi() {
  int now = n;
  int val = 0;
  int num = 1;
  int now_num = 0;
  int res = 0;
  while (now--) {
    res += val;
    now_num++;
    if (now_num == num) {
      num = num * 2;
      now_num = 0;
      val++;
    }
  }
  return res;
}
int get_mx() { return (0 + n - 1) * n / 2; }
int mi, mx;
const int maxn = 5005;
int f[maxn];
void init() {
  for (int i = 0; i < n; i++) {
    f[i] = 1;
  }
  int now = n - 1;
  int now_push = 1;
  int val = mx;
  while (true) {
    int tmp = val - (now - now_push);
    if (d > tmp) {
      break;
    }
    f[now]--;
    f[now_push]++;
    if (f[now_push] == (1 << now_push)) {
      now_push++;
    }
    val = tmp;
    now--;
    if (val == mi) {
      break;
    }
  }
  int tmp = val - d;
  f[now]--;
  f[now - tmp]++;
}
vector<int> V2[maxn];
int ans[maxn];
void out() {
  for (int i = 0; i < n; i++) {
    V2[i].clear();
  }
  int now = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < f[i]; j++) {
      V2[i].push_back(now);
      now++;
    }
  }
  for (int i = 1; i < n; i++) {
    for (int j = 0; j < V2[i].size(); j++) {
      ans[V2[i][j]] = V2[i - 1][j / 2];
    }
  }
  puts("YES");
  for (int i = 2; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  puts("");
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d", &n, &d);
    mi = get_mi();
    mx = get_mx();
    if (d < mi || d > mx) {
      puts("NO");
      continue;
    }
    if (n == 2) {
      puts("YES");
      printf("1\n");
      continue;
    }
    init();
    out();
  }
}