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;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
const int MOD = 1e9 + 7;
const double eps = 1e-5;
const int INF = 0x3f3f3f3f;
const long long INFLL = 0x7f7f7f7f;
int gcd(int a, int b) { return (a % b == 0) ? b : gcd(b, a % b); }
long long lowbit(long long x) { return x & -x; }
const int N1 = 1005;
const int N = 2e5 + 5;
int cnt[5005];
int minn[5005], maxx[5005];
int can[55];
int ans[5005];
int flag[5005];
queue<int> q[5005];
int main() {
  int t;
  scanf("%d", &t);
  int now = 2, num = 0, p = 1;
  for (int i = 2; i <= 5000; i++) {
    minn[i] = minn[i - 1] + p;
    num++;
    if (num == now) {
      now *= 2;
      num = 0;
      p++;
    }
  }
  for (int i = 2; i <= 5000; ++i) maxx[i] = maxx[i - 1] + (i - 1);
  can[1] = 2;
  for (int i = 2; i < 20; ++i) can[i] = can[i - 1] * 2;
  while (t--) {
    int n, d;
    scanf("%d%d", &n, &d);
    if (d < minn[n] || d > maxx[n]) {
      printf("NO\n");
      continue;
    }
    for (int i = 1; i <= n; i++)
      while (!q[i].empty()) q[i].pop();
    for (int i = 1; i < n; i++) {
      cnt[i] = 1;
      q[i].push(i + 1);
      ans[i + 1] = i;
      flag[i] = 0;
    }
    int sum = maxx[n];
    int da = n - 1, xiao = 1;
    while (sum > d) {
      if (sum - d >= da - xiao) {
        sum -= (da - xiao);
        cnt[da]--;
        cnt[xiao]++;
        q[xiao].push(q[da].front());
        ans[q[da].front()] = xiao;
        q[da].pop();
        da--;
        if (cnt[xiao] == can[xiao]) xiao++;
      } else {
        cnt[da]--;
        cnt[da - (sum - d)]++;
        q[da - (sum - d)].push(q[da].front());
        ans[q[da].front()] = da - (sum - d);
        q[da].pop();
        sum = d;
      }
    }
    printf("YES\n");
    q[0].push(1);
    for (int i = 2; i <= n; i++) {
      printf("%d ", q[ans[i] - 1].front());
      flag[ans[i] - 1]++;
      if (flag[ans[i] - 1] == 2) {
        flag[ans[i] - 1] = 0;
        q[ans[i] - 1].pop();
      }
    }
    printf("\n");
  }
  return 0;
}