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 INF = 0x3f3f3f3f;
const double pi = acos(-1), eps = 1e-10;
const int maxn = 1 << 17;
const int N = 5e3 + 10, M = N * 40;
int n, d;
int fa[N];
int son[N];
int de[N];
int dp[N];
int main() {
  de[1] = 1;
  int t;
  cin >> t;
  while (t--) {
    scanf("%d%d", &n, &d);
    memset(de, 0, sizeof(de));
    de[1] = 1;
    int s = 0;
    for (int i = 1; i <= n; i++) s += i - 1;
    int now = 1;
    for (int i = 1; i <= n; i++) {
      dp[i] = dp[i - 1] + 1;
      de[dp[i]]--;
      if (s - (n - i + 1) >= d) {
        if (de[dp[i] - 1]) {
          s -= (n - i + 1);
          de[dp[i] - 1]--;
          de[dp[i]]++;
          dp[i]--;
        }
      }
      de[dp[i] + 1] += 2;
    }
    if (s != d) {
      printf("NO\n");
      continue;
    }
    int flag = 1;
    memset(son, 0, sizeof(son));
    for (int i = 2; i <= n; i++) {
      fa[i] = 0;
      for (int j = 1; j < i; j++) {
        if (son[j] < 2 && dp[i] == dp[j] + 1) {
          son[j]++;
          fa[i] = j;
          break;
        }
      }
      if (fa[i] == 0) {
        flag = 0;
        break;
      }
    }
    if (flag == 0) {
      printf("NO\n");
    } else
      printf("YES\n");
    for (int i = 2; i <= n; i++) printf("%d ", fa[i]);
    printf("\n");
  }
  return 0;
}