#include <bits/stdc++.h>
using namespace std;
const int N = 5007, inf = 0x3f3f3f3f;
int men[N], mai[N];
int p[N];
void solve(int u, int d, int n) {
  assert(d >= men[n] and d <= mai[n]);
  if (n <= 1) return;
  d -= n - 1;
  for (int j = 0; j <= n - 1; ++j) {
    int men2 = men[j] + men[n - 1 - j];
    int mai2 = mai[j] + mai[n - 1 - j];
    if (d >= men2 and d <= mai2) {
      int v = u + 1;
      int d1 = min(mai[j], d - men[n - 1 - j]);
      int d2 = d - d1;
      p[v] = u;
      solve(v, d1, j);
      v = u + 1 + j;
      p[v] = u;
      solve(v, d2, n - 1 - j);
      return;
    }
  }
  assert(0);
}
int main() {
  int t;
  cin >> t;
  int d = -1;
  for (int i = 1; i <= 5000; ++i) {
    if (__builtin_popcount(i) == 1) d++;
    men[i] = men[i - 1] + d;
    mai[i] = mai[i - 1] + (i - 1);
  }
  while (t--) {
    int n, d;
    cin >> n >> d;
    if (d < men[n] or d > mai[n]) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      solve(1, d, n);
      for (int i = 2; i <= n; ++i) cout << p[i] << ' ';
      cout << '\n';
    }
  }
}

### 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;
const int N = 5007, inf = 0x3f3f3f3f;
int men[N], mai[N];
int p[N];
void solve(int u, int d, int n) {
  assert(d >= men[n] and d <= mai[n]);
  if (n <= 1) return;
  d -= n - 1;
  for (int j = 0; j <= n - 1; ++j) {
    int men2 = men[j] + men[n - 1 - j];
    int mai2 = mai[j] + mai[n - 1 - j];
    if (d >= men2 and d <= mai2) {
      int v = u + 1;
      int d1 = min(mai[j], d - men[n - 1 - j]);
      int d2 = d - d1;
      p[v] = u;
      solve(v, d1, j);
      v = u + 1 + j;
      p[v] = u;
      solve(v, d2, n - 1 - j);
      return;
    }
  }
  assert(0);
}
int main() {
  int t;
  cin >> t;
  int d = -1;
  for (int i = 1; i <= 5000; ++i) {
    if (__builtin_popcount(i) == 1) d++;
    men[i] = men[i - 1] + d;
    mai[i] = mai[i - 1] + (i - 1);
  }
  while (t--) {
    int n, d;
    cin >> n >> d;
    if (d < men[n] or d > mai[n]) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      solve(1, d, n);
      for (int i = 2; i <= n; ++i) cout << p[i] << ' ';
      cout << '\n';
    }
  }
}

```