#include <bits/stdc++.h>
using namespace std;
int n, d;
int pa[5020];
int ch[5020];
int lg2[8200];
bool mi[5020];
int dep[8200];
void getlog2() {
  lg2[1] = 0;
  for (int i = 1; i <= 13; i++)
    for (int j = (1 << i); j < (1 << (i + 1)); j++) {
      lg2[j] = i;
    }
}
void solve() {
  cin >> n >> d;
  int ub = n * (n - 1) / 2, lb = 0;
  for (int i = 1; i <= n; i++) {
    pa[i] = i - 1;
    lb += lg2[i];
    mi[i] = false;
    dep[i] = i - 1;
  }
  for (int i = 1; i < n; i++) ch[i] = 1;
  ch[n] = 0;
  if (d < lb || d > ub) {
    cout << "NO" << endl;
    return;
  }
  while (ub > d) {
    int v = -1;
    for (int i = 1; i <= n; i++)
      if (ch[i] == 0 && !mi[i] && (v == -1 || dep[i] < dep[v])) {
        v = i;
      }
    int p = -1;
    for (int i = 1; i <= n; i++)
      if (ch[i] < 2 && dep[i] + 2 <= dep[v] && (p == -1 || dep[i] > dep[p])) {
        p = i;
      }
    if (p == -1) {
      mi[v] = true;
      continue;
    }
    ch[p]++;
    ch[pa[v]]--;
    pa[v] = p;
    dep[v]--;
    ub--;
  }
  cout << "YES" << endl;
  for (int i = 2; i <= n; i++) cout << pa[i] << ' ';
  cout << endl;
}
int main() {
  getlog2();
  int testnum;
  cin >> testnum;
  while (testnum--) solve();
  return 0;
}

### Prompt

Your task is to create 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;
int pa[5020];
int ch[5020];
int lg2[8200];
bool mi[5020];
int dep[8200];
void getlog2() {
  lg2[1] = 0;
  for (int i = 1; i <= 13; i++)
    for (int j = (1 << i); j < (1 << (i + 1)); j++) {
      lg2[j] = i;
    }
}
void solve() {
  cin >> n >> d;
  int ub = n * (n - 1) / 2, lb = 0;
  for (int i = 1; i <= n; i++) {
    pa[i] = i - 1;
    lb += lg2[i];
    mi[i] = false;
    dep[i] = i - 1;
  }
  for (int i = 1; i < n; i++) ch[i] = 1;
  ch[n] = 0;
  if (d < lb || d > ub) {
    cout << "NO" << endl;
    return;
  }
  while (ub > d) {
    int v = -1;
    for (int i = 1; i <= n; i++)
      if (ch[i] == 0 && !mi[i] && (v == -1 || dep[i] < dep[v])) {
        v = i;
      }
    int p = -1;
    for (int i = 1; i <= n; i++)
      if (ch[i] < 2 && dep[i] + 2 <= dep[v] && (p == -1 || dep[i] > dep[p])) {
        p = i;
      }
    if (p == -1) {
      mi[v] = true;
      continue;
    }
    ch[p]++;
    ch[pa[v]]--;
    pa[v] = p;
    dep[v]--;
    ub--;
  }
  cout << "YES" << endl;
  for (int i = 2; i <= n; i++) cout << pa[i] << ' ';
  cout << endl;
}
int main() {
  getlog2();
  int testnum;
  cin >> testnum;
  while (testnum--) solve();
  return 0;
}

```