#include <bits/stdc++.h>
using namespace std;
int n, d, p[5010], ch[5010], bad[5010], lv[5010];
int main(void) {
  int q;
  scanf("%d", &q);
  while (q--) {
    scanf("%d %d", &n, &d);
    int lm = 0;
    for (int i = 1, dp = -1; i <= n; i++) {
      if ((i & (i - 1)) == 0) dp++;
      lm += dp;
    }
    int rm = n * (n - 1) / 2;
    if (d < lm || rm < d) {
      printf("NO\n");
    } else {
      for (int i = 2; i <= n; i++) {
        p[i] = i - 1;
        lv[i] = lv[i - 1] + 1;
        ch[i - 1] = 1;
        bad[i] = false;
      }
      ch[n] = 0;
      while (d < rm) {
        int v = 0;
        lv[v] = 1e9;
        for (int i = 1; i <= n; i++) {
          if (!ch[i] && !bad[i] && lv[i] < lv[v]) {
            v = i;
          }
        }
        assert(v != 0);
        int ok = false;
        for (int i = 1; i <= n; i++) {
          if (ch[i] != 2 && lv[i] + 2 == lv[v]) {
            ch[p[v]]--;
            ch[i]++;
            p[v] = i;
            lv[v] = lv[i] + 1;
            rm--;
            ok = true;
            break;
          }
        }
        if (!ok) bad[v] = true;
      }
      printf("YES\n");
      for (int i = 2; i <= n; i++) {
        printf("%d ", p[i]);
      }
      puts("");
    }
  }
}

### Prompt

Please formulate a Cpp solution to the following 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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, d, p[5010], ch[5010], bad[5010], lv[5010];
int main(void) {
  int q;
  scanf("%d", &q);
  while (q--) {
    scanf("%d %d", &n, &d);
    int lm = 0;
    for (int i = 1, dp = -1; i <= n; i++) {
      if ((i & (i - 1)) == 0) dp++;
      lm += dp;
    }
    int rm = n * (n - 1) / 2;
    if (d < lm || rm < d) {
      printf("NO\n");
    } else {
      for (int i = 2; i <= n; i++) {
        p[i] = i - 1;
        lv[i] = lv[i - 1] + 1;
        ch[i - 1] = 1;
        bad[i] = false;
      }
      ch[n] = 0;
      while (d < rm) {
        int v = 0;
        lv[v] = 1e9;
        for (int i = 1; i <= n; i++) {
          if (!ch[i] && !bad[i] && lv[i] < lv[v]) {
            v = i;
          }
        }
        assert(v != 0);
        int ok = false;
        for (int i = 1; i <= n; i++) {
          if (ch[i] != 2 && lv[i] + 2 == lv[v]) {
            ch[p[v]]--;
            ch[i]++;
            p[v] = i;
            lv[v] = lv[i] + 1;
            rm--;
            ok = true;
            break;
          }
        }
        if (!ok) bad[v] = true;
      }
      printf("YES\n");
      for (int i = 2; i <= n; i++) {
        printf("%d ", p[i]);
      }
      puts("");
    }
  }
}

```