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:

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<image>
#include <bits/stdc++.h>
using namespace std;
const int maxn = 5000;
int mi[5001], mx[5001];
int a[maxn + 5];
int n, m;
vector<int> tmp;
void solve() {
  if (m < mi[n] || m > mx[n]) {
    printf("NO\n");
    return;
  }
  printf("YES\n");
  int now = n;
  for (int i = 0; i <= n; ++i) a[i] = 0;
  int d;
  for (d = 0; now >= (1 << d); ++d) a[d] = 1 << d, now -= (1 << d);
  a[d] = now;
  m -= mi[n];
  while (m--) {
    for (int i = d; i >= 0; --i)
      if (a[i + 1] + 1 <= 2 * (a[i] - 1)) {
        ++a[i + 1];
        --a[i];
        if (i == d) ++d;
        break;
      }
  }
  tmp.clear();
  tmp.push_back(1);
  tmp.push_back(1);
  int id = 2;
  for (int i = 1; i <= d; ++i) {
    for (int j = 0; j < a[i]; ++j) printf("%d ", tmp[j]);
    tmp.clear();
    for (int j = id; j < id + a[i]; ++j) tmp.push_back(j), tmp.push_back(j);
    id += a[i];
  }
  printf("\n");
}
int main() {
  mi[1] = 0;
  mx[1] = 0;
  int dep = 1;
  for (int i = 2; i <= 5000; ++i) {
    mi[i] = mi[i - 1] + dep;
    if (i + 1 == (1 << (dep + 1))) ++dep;
    mx[i] = mx[i - 1] + i - 1;
  }
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n >> m;
    solve();
  }
}
