#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const double eps = 1e-8;
const int INF32 = 0x3f3f3f3f;
const long long INF64 = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const unsigned long long mod2 = 1;
const int N = 5e3 + 5;
const int sz = (1 << 20);
int num[N], ans[N], cnt;
void dfs(int dpt, int fa) {
  ans[++cnt] = fa;
  int u = cnt;
  if (num[dpt]) {
    --num[dpt];
    dfs(dpt + 1, u);
  }
  if (num[dpt]) {
    --num[dpt];
    dfs(dpt + 1, u);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int q;
  cin >> q;
  while (q--) {
    int n, d;
    cin >> n >> d;
    for (int i = 0; i < n; ++i) {
      num[i] = 1;
      d -= i;
    }
    int l = 1, r = n - 1;
    while (d) {
      if (l - r < d) {
        ++l;
      } else {
        d += r - l;
        ++num[l];
        --num[r];
        --r;
        if (num[l] == (1 << min(l, 20))) ++l;
      }
      if (l >= r) break;
    }
    if (l >= r && d) {
      cout << "NO" << '\n';
      continue;
    }
    cout << "YES" << '\n';
    cnt = 0;
    dfs(1, 0);
    for (int i = 2; i <= n; ++i) {
      cout << ans[i] << ' ';
    }
    cout << '\n';
  }
  return 0;
}

### 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const double eps = 1e-8;
const int INF32 = 0x3f3f3f3f;
const long long INF64 = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const unsigned long long mod2 = 1;
const int N = 5e3 + 5;
const int sz = (1 << 20);
int num[N], ans[N], cnt;
void dfs(int dpt, int fa) {
  ans[++cnt] = fa;
  int u = cnt;
  if (num[dpt]) {
    --num[dpt];
    dfs(dpt + 1, u);
  }
  if (num[dpt]) {
    --num[dpt];
    dfs(dpt + 1, u);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int q;
  cin >> q;
  while (q--) {
    int n, d;
    cin >> n >> d;
    for (int i = 0; i < n; ++i) {
      num[i] = 1;
      d -= i;
    }
    int l = 1, r = n - 1;
    while (d) {
      if (l - r < d) {
        ++l;
      } else {
        d += r - l;
        ++num[l];
        --num[r];
        --r;
        if (num[l] == (1 << min(l, 20))) ++l;
      }
      if (l >= r) break;
    }
    if (l >= r && d) {
      cout << "NO" << '\n';
      continue;
    }
    cout << "YES" << '\n';
    cnt = 0;
    dfs(1, 0);
    for (int i = 2; i <= n; ++i) {
      cout << ans[i] << ' ';
    }
    cout << '\n';
  }
  return 0;
}

```