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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5e3 + 10;
const int MOD = 998244353;
const int INF = 0x3f3f3f3f;
int N, M;
int num[20];
int ver[20];
int each[20];
int fa[MAXN];
void mark() {
  for (int i = 0; i < 20; i++) {
    each[i] = pow(2, i);
    num[i] = i * each[i];
  }
  ver[0] = each[0];
  for (int i = 1; i < 20; i++) {
    ver[i] = ver[i - 1] + each[i];
    num[i] += num[i - 1];
  }
}
void init(int x) {
  queue<int> q;
  for (int i = 2; i <= x; i++) {
    fa[i] = i - 1;
  }
  while (!q.empty()) {
    q.pop();
  }
  x++;
  for (int i = 0; x <= N; i++) {
    q.push(i + 1);
    for (int j = 1; j < each[i] && x <= N; j++, x++) {
      fa[x] = q.front();
      q.pop();
      q.push(x);
      q.push(x);
    }
  }
}
void solve(int x) {
  int now;
  int det;
  int floor;
  x++;
  init(x);
  now = N;
  floor = x;
  for (int i = x - 1; i > 0 && M; i--) {
    for (int j = now - ver[i - 1] - floor + i; j > 0 && M; j--, now--) {
      det = N - now + floor - i;
      if (det <= M) {
        M -= det;
        fa[now] = x;
        x = now;
      } else {
        det -= M;
        M = 0;
        while (det--) {
          x = fa[x];
        }
        fa[now] = x;
      }
    }
  }
}
int main() {
  int T;
  int mini;
  int maxi;
  int x;
  mark();
  while (~scanf("%d", &T)) {
    while (T--) {
      scanf("%d%d", &N, &M);
      for (int i = 0; i < 20; i++) {
        if (N <= ver[i]) {
          mini = num[i - 1] + (N - ver[i - 1]) * i;
          x = i;
          break;
        } else if (N == ver[i]) {
          x = i;
          break;
        }
      }
      maxi = (N - 1) * (N) / 2;
      if (M < mini || M > maxi) {
        printf("NO\n");
        continue;
      }
      printf("YES\n");
      M -= mini;
      solve(x);
      for (int i = 2; i <= N; i++) {
        printf("%d%c", fa[i], i == N ? '\n' : ' ');
      }
    }
  }
  return 0;
}