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 + 5;
int cnt[MAXN], son[MAXN], fa[MAXN];
vector<int> E[MAXN];
void solve() {
  int n, d, sum = 0;
  scanf("%d%d", &n, &d);
  for (int i = 0; i < n; i++) {
    cnt[i] = 0;
    son[i + 1] = 0;
    E[i].clear();
    cnt[i]++;
    sum += (i);
  }
  if (sum < d) {
    puts("NO");
    return;
  }
  int l = 1, r = n - 1;
  bool flag = true;
  while (sum > d && flag) {
    int det = sum - d;
    if (l >= r) {
      flag = false;
      break;
    }
    if (det > r - l) {
      sum -= r - l;
      cnt[r]--;
      r--;
      cnt[l]++;
    } else {
      cnt[r - det]++;
      cnt[r]--;
      sum -= det;
    }
    if (cnt[l] == (1 << l)) l++;
  }
  if (!flag)
    puts("NO");
  else {
    puts("YES");
    E[0].push_back(1);
    int now = 2;
    for (int i = 1; i < n; i++) {
      for (int j = 0; j < cnt[i]; j++) {
        int f = *prev(E[i - 1].end());
        son[f]++;
        fa[now] = f;
        E[i].push_back(now);
        if (son[f] == 2) E[i - 1].pop_back();
        now++;
      }
    }
    assert(now == n + 1);
    for (int i = 2; i <= n; i++) printf("%d ", fa[i]);
    puts("");
  }
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    solve();
  }
  return 0;
}
