#include <bits/stdc++.h>
using namespace std;
const int N = 5000 + 50;
int p[N];
int mi[N], ma[N];
int calc_max(int n) { return n * (n - 1) / 2; }
int calc_min(int n) {
  int res = 0;
  int d = 0, cnt = 1;
  while (n > cnt) {
    n -= cnt;
    res += cnt * d;
    d++;
    cnt <<= 1;
  }
  res += n * d;
  return res;
}
bool can(int n, int d) {
  if (d < calc_min(n) || d > calc_max(n)) return false;
  return true;
}
bool dfs(int L, int R, int d) {
  int n = R - L + 1;
  if (n == 1) return true;
  if (!can(n, d)) return false;
  d -= n - 1;
  if (can(n - 1, d)) {
    p[L + 1] = L;
    return dfs(L + 1, R, d);
  }
  for (int i = 1; i < n - 1; i++) {
    for (int j = mi[i]; j <= ma[i] && j <= d; j++) {
      if (can(n - 1 - i, d - j)) {
        p[L + 1] = L;
        dfs(L + 1, L + i, j);
        p[L + i + 1] = L;
        dfs(L + i + 1, R, d - j);
        return true;
      }
    }
  }
  return false;
}
int main() {
  for (int i = 0; i < N; i++) {
    mi[i] = calc_min(i);
    ma[i] = calc_max(i);
  }
  ios::sync_with_stdio(false);
  int T;
  cin >> T;
  while (T--) {
    int n, d;
    cin >> n >> d;
    bool ok = dfs(1, n, d);
    if (!ok)
      cout << "NO" << endl;
    else {
      cout << "YES" << endl;
      for (int i = 2; i <= n; i++) {
        cout << p[i] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
const int N = 5000 + 50;
int p[N];
int mi[N], ma[N];
int calc_max(int n) { return n * (n - 1) / 2; }
int calc_min(int n) {
  int res = 0;
  int d = 0, cnt = 1;
  while (n > cnt) {
    n -= cnt;
    res += cnt * d;
    d++;
    cnt <<= 1;
  }
  res += n * d;
  return res;
}
bool can(int n, int d) {
  if (d < calc_min(n) || d > calc_max(n)) return false;
  return true;
}
bool dfs(int L, int R, int d) {
  int n = R - L + 1;
  if (n == 1) return true;
  if (!can(n, d)) return false;
  d -= n - 1;
  if (can(n - 1, d)) {
    p[L + 1] = L;
    return dfs(L + 1, R, d);
  }
  for (int i = 1; i < n - 1; i++) {
    for (int j = mi[i]; j <= ma[i] && j <= d; j++) {
      if (can(n - 1 - i, d - j)) {
        p[L + 1] = L;
        dfs(L + 1, L + i, j);
        p[L + i + 1] = L;
        dfs(L + i + 1, R, d - j);
        return true;
      }
    }
  }
  return false;
}
int main() {
  for (int i = 0; i < N; i++) {
    mi[i] = calc_min(i);
    ma[i] = calc_max(i);
  }
  ios::sync_with_stdio(false);
  int T;
  cin >> T;
  while (T--) {
    int n, d;
    cin >> n >> d;
    bool ok = dfs(1, n, d);
    if (!ok)
      cout << "NO" << endl;
    else {
      cout << "YES" << endl;
      for (int i = 2; i <= n; i++) {
        cout << p[i] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}

```