#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    int ans = n * (n - 1) / 2;
    if (d > ans) {
      puts("NO");
      continue;
    }
    int father[n + 1];
    int son[n + 1];
    int deep[n + 1];
    int bad[n + 1];
    deep[1] = 0;
    deep[0] = 1000000;
    bad[0] = 1;
    bad[1] = 1;
    for (int i = 2; i <= n; i++) {
      son[i] = 0;
      father[i] = i - 1;
      son[i - 1] = 1;
      deep[i] = deep[i - 1] + 1;
      bad[i] = 0;
    }
    while (ans > d) {
      int p = 0;
      for (int i = 1; i <= n; i++) {
        if (son[i] == 0 && !bad[i]) {
          if (deep[i] < deep[p]) {
            p = i;
          }
        }
      }
      if (p == 0) break;
      int j;
      int k = 0;
      int m = 0;
      int aux = 0;
      for (j = 1; j <= n; j++) {
        aux = (deep[j] + 1 - deep[p]);
        if (j != p && son[j] < 2 && aux < 0 && ans + aux >= d) {
          if (aux < m) {
            k = j;
            m = aux;
          }
        }
      }
      if (!m) {
        bad[p] = 1;
        continue;
      }
      son[father[p]]--;
      father[p] = k;
      son[k]++;
      deep[p] = deep[k] + 1;
      ans += m;
    }
    if (ans == d) {
      puts("YES");
      for (int i = 2; i <= n; i++) {
        printf("%d ", father[i]);
      }
      puts("");
    } else
      puts("NO");
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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 main() {
  int t;
  cin >> t;
  while (t--) {
    int n, d;
    cin >> n >> d;
    int ans = n * (n - 1) / 2;
    if (d > ans) {
      puts("NO");
      continue;
    }
    int father[n + 1];
    int son[n + 1];
    int deep[n + 1];
    int bad[n + 1];
    deep[1] = 0;
    deep[0] = 1000000;
    bad[0] = 1;
    bad[1] = 1;
    for (int i = 2; i <= n; i++) {
      son[i] = 0;
      father[i] = i - 1;
      son[i - 1] = 1;
      deep[i] = deep[i - 1] + 1;
      bad[i] = 0;
    }
    while (ans > d) {
      int p = 0;
      for (int i = 1; i <= n; i++) {
        if (son[i] == 0 && !bad[i]) {
          if (deep[i] < deep[p]) {
            p = i;
          }
        }
      }
      if (p == 0) break;
      int j;
      int k = 0;
      int m = 0;
      int aux = 0;
      for (j = 1; j <= n; j++) {
        aux = (deep[j] + 1 - deep[p]);
        if (j != p && son[j] < 2 && aux < 0 && ans + aux >= d) {
          if (aux < m) {
            k = j;
            m = aux;
          }
        }
      }
      if (!m) {
        bad[p] = 1;
        continue;
      }
      son[father[p]]--;
      father[p] = k;
      son[k]++;
      deep[p] = deep[k] + 1;
      ans += m;
    }
    if (ans == d) {
      puts("YES");
      for (int i = 2; i <= n; i++) {
        printf("%d ", father[i]);
      }
      puts("");
    } else
      puts("NO");
  }
}

```