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>
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int ret = 0, flag = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') flag = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') ret = ret * 10 + (c ^ '0'), c = getchar();
  return flag * ret;
}
struct use {
  int x, dis;
};
const int N = 5e3 + 5;
int T, n, d;
vector<int> edge[N];
int dad[N];
int build(int x, int dis) {
  int sum = 0;
  if ((x << 1) <= n) {
    edge[x].push_back(x << 1);
    sum += dis + 1;
    sum += build(x << 1, dis + 1);
  }
  if ((x << 1 | 1) <= n) {
    edge[x].push_back(x << 1 | 1);
    sum += dis + 1;
    sum += build(x << 1 | 1, dis + 1);
  }
  return sum;
}
void bfs() {
  queue<use> q;
  q.push({1, 0});
  int now = 0;
  vector<int> p1, p2;
  while (q.empty() == 0) {
    int x = q.front().x, dis = q.front().dis;
    q.pop();
    if (now != dis) {
      now = dis, p1.clear(), p2.clear();
    }
    for (auto it = edge[x].begin(); it != edge[x].end(); it++) {
      int y = *it;
      if (edge[y].size() == 0) {
        p1.push_back(y);
        if (p2.size()) {
          edge[p2[0]].push_back(y);
          edge[x].erase(it);
          return;
        }
      }
      if (edge[y].size() != 2) p2.push_back(y);
      q.push({y, dis + 1});
    }
  }
}
int main() {
  T = read();
  while (T--) {
    n = read(), d = read();
    for (int i = 1; i <= n; i++) edge[i].clear();
    int mn = build(1, 0);
    int mx = (n - 1) * n / 2;
    if (!(mn <= d && d <= mx)) {
      printf("NO\n");
      continue;
    }
    int now = mn;
    while (now < d) {
      bfs();
      now++;
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < edge[i].size(); j++) dad[edge[i][j]] = i;
    }
    printf("YES\n");
    for (int i = 2; i <= n; i++) printf("%d ", dad[i]);
    printf("\n");
  }
  return 0;
}
