#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, inf = 1e9 + 5;
int n, m;
int d[N], a[N], fa[N];
int main() {
  int t, ans;
  scanf("%d", &t);
  while (t--) {
    scanf("%d %d", &n, &m);
    int sum = n * (n - 1) / 2 - m, l = 2, r = n;
    for (int i = 1; i < n + 1; ++i) d[i] = 1;
    while (sum > 0) {
      while (l < r && (sum - r + l < 0 || d[l] == 1 << (l - 1))) l++;
      if (l >= r) break;
      sum -= r - l;
      d[l] += 1;
      d[r] -= 1;
      if (d[r] == 0) r -= 1;
    }
    if (sum != 0) {
      puts("NO");
      continue;
    }
    puts("YES");
    for (int i = 1; i < n + 1; ++i) a[i] = d[i] + a[i - 1];
    for (int i = 2; i < n + 1; ++i) {
      if (d[i] == 0) break;
      int x = d[i], id = a[i - 1] + 1;
      for (int j = a[i - 2] + 1; j < a[i - 1] + 1; ++j) {
        fa[id] = j;
        --x;
        ++id;
        if (!x) break;
        fa[id] = j;
        --x;
        ++id;
        if (!x) break;
      }
    }
    for (int i = 2; i < n + 1; ++i) printf("%d ", fa[i]);
    puts("");
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
const int N = 2e5 + 5, inf = 1e9 + 5;
int n, m;
int d[N], a[N], fa[N];
int main() {
  int t, ans;
  scanf("%d", &t);
  while (t--) {
    scanf("%d %d", &n, &m);
    int sum = n * (n - 1) / 2 - m, l = 2, r = n;
    for (int i = 1; i < n + 1; ++i) d[i] = 1;
    while (sum > 0) {
      while (l < r && (sum - r + l < 0 || d[l] == 1 << (l - 1))) l++;
      if (l >= r) break;
      sum -= r - l;
      d[l] += 1;
      d[r] -= 1;
      if (d[r] == 0) r -= 1;
    }
    if (sum != 0) {
      puts("NO");
      continue;
    }
    puts("YES");
    for (int i = 1; i < n + 1; ++i) a[i] = d[i] + a[i - 1];
    for (int i = 2; i < n + 1; ++i) {
      if (d[i] == 0) break;
      int x = d[i], id = a[i - 1] + 1;
      for (int j = a[i - 2] + 1; j < a[i - 1] + 1; ++j) {
        fa[id] = j;
        --x;
        ++id;
        if (!x) break;
        fa[id] = j;
        --x;
        ++id;
        if (!x) break;
      }
    }
    for (int i = 2; i < n + 1; ++i) printf("%d ", fa[i]);
    puts("");
  }
  return 0;
}

```