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;
int Gcd(int a, int b) {
  if (b == 0) return a;
  return Gcd(b, a % b);
}
int Lcm(int a, int b) { return a / Gcd(a, b) * b; }
inline long long read() {
  long long f = 1, x = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int maxn = 1e6 + 10;
int dep[maxn];
int main() {
  int T = read();
  while (T--) {
    int n = read(), d = read(), sum = 0;
    for (int i = 1; i <= n; i++) dep[i] = 0;
    for (int i = 1; i <= n; i++) {
      sum += i - 1;
      dep[i]++;
    }
    if (sum < d) {
      puts("NO");
      continue;
    }
    for (int i = n; i >= 3; i--) {
      if (sum == d) break;
      int tmp = i;
      if (dep[i]) {
        while (sum > d && dep[tmp - 1] < dep[tmp - 2] * 2) {
          dep[tmp]--;
          dep[tmp - 1]++;
          tmp--;
          sum--;
        }
      }
    }
    if (sum == d) {
      int id = 1;
      vector<int> depth(n + 1, 0);
      vector<int> son(n + 1, 0), fa(n + 1, 0);
      for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= dep[i]; j++) {
          depth[id++] = i;
        }
      }
      int tag = 1;
      for (int i = 2; i <= n; i++) {
        int res = 0;
        for (int j = 1; j < i; j++) {
          if (depth[j] == depth[i] - 1 && son[j] < 2) {
            son[j]++;
            fa[i] = j;
            res = 1;
            break;
          }
        }
        if (!res) {
          tag = 0;
        }
      }
      if (!tag)
        puts("NO");
      else {
        puts("YES");
        for (int i = 2; i <= n; i++) {
          printf("%d ", fa[i]);
        }
        printf("\n");
      }
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}