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 N = 5111;
int n, d;
int deg[N], par[N], depth[N];
bool check[N];
void solve() {
  cin >> n >> d;
  for (int i = 2; i <= n; ++i) {
    depth[i] = i - 1;
    par[i] = i - 1;
  }
  for (int i = 1; i < n; ++i) deg[i] = 1;
  deg[n] = 0;
  for (int i = 1; i <= n; ++i) check[i] = 0;
  int sum = n * (n - 1) / 2;
  while (sum > d) {
    int v = -1;
    for (int i = 1; i <= n; ++i) {
      if (deg[i] == 0 && !check[i]) {
        if (v == -1 || depth[i] < depth[v]) v = i;
      }
    }
    if (v == -1) break;
    int u = -1;
    for (int i = 1; i <= n; ++i) {
      if (depth[i] < depth[par[v]] && deg[i] < 2 &&
          sum - depth[v] + depth[i] + 1 >= d) {
        if (u == -1 || depth[i] < depth[u]) u = i;
      }
    }
    if (u == -1) {
      check[v] = 1;
      continue;
    }
    sum = sum - depth[v] + depth[u] + 1;
    deg[par[v]]--, deg[u]++;
    par[v] = u;
    depth[v] = depth[u] + 1;
  }
  if (sum != d) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  for (int i = 2; i <= n; ++i) cout << par[i] << " ";
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int numTest;
  cin >> numTest;
  while (numTest--) solve();
  return 0;
}
