#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5010;
int f[MAXN], ch[MAXN], dep[MAXN];
bool vis[MAXN];
int main() {
  int n, d;
  int T;
  scanf("%d", &T);
  while (T--) {
    memset(vis, 0, sizeof(vis));
    memset(ch, 0, sizeof(ch));
    scanf("%d%d", &n, &d);
    int INF = (n * (n - 1)) / 2;
    int cnt_ = 0, cnt = 0;
    for (int i = 1; i <= n; i *= 2) {
      for (int j = i; j < i * 2 && j <= n; j++) {
        cnt += cnt_;
      }
      cnt_++;
    }
    if (d < cnt || d > INF) {
      printf("NO\n");
      continue;
    }
    int cur = INF;
    for (int i = 1; i <= n; i++) dep[i] = i - 1, f[i] = i - 1, ch[i] = 1;
    ch[n] = 0;
    while (cur > d) {
      int v = -1;
      for (int i = 1; i <= n; i++) {
        if (ch[i] == 0 && vis[i] == 0 && (v == -1 || dep[v] > dep[i])) v = i;
      }
      if (v == -1) break;
      int fa = -1;
      for (int i = 1; i <= n; i++) {
        if (ch[i] < 2 && dep[i] < dep[v] - 1 && (fa == -1 || dep[fa] < dep[i]))
          fa = i;
      }
      if (fa == -1) {
        vis[v] = 1;
        continue;
      }
      if (dep[v] - dep[fa] != 2) break;
      ch[f[v]]--;
      ch[fa]++;
      dep[v]--;
      f[v] = fa;
      cur--;
    }
    printf("YES\n");
    for (int i = 2; i <= n; i++) {
      printf("%d%c", f[i], i == n ? '\n' : ' ');
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5010;
int f[MAXN], ch[MAXN], dep[MAXN];
bool vis[MAXN];
int main() {
  int n, d;
  int T;
  scanf("%d", &T);
  while (T--) {
    memset(vis, 0, sizeof(vis));
    memset(ch, 0, sizeof(ch));
    scanf("%d%d", &n, &d);
    int INF = (n * (n - 1)) / 2;
    int cnt_ = 0, cnt = 0;
    for (int i = 1; i <= n; i *= 2) {
      for (int j = i; j < i * 2 && j <= n; j++) {
        cnt += cnt_;
      }
      cnt_++;
    }
    if (d < cnt || d > INF) {
      printf("NO\n");
      continue;
    }
    int cur = INF;
    for (int i = 1; i <= n; i++) dep[i] = i - 1, f[i] = i - 1, ch[i] = 1;
    ch[n] = 0;
    while (cur > d) {
      int v = -1;
      for (int i = 1; i <= n; i++) {
        if (ch[i] == 0 && vis[i] == 0 && (v == -1 || dep[v] > dep[i])) v = i;
      }
      if (v == -1) break;
      int fa = -1;
      for (int i = 1; i <= n; i++) {
        if (ch[i] < 2 && dep[i] < dep[v] - 1 && (fa == -1 || dep[fa] < dep[i]))
          fa = i;
      }
      if (fa == -1) {
        vis[v] = 1;
        continue;
      }
      if (dep[v] - dep[fa] != 2) break;
      ch[f[v]]--;
      ch[fa]++;
      dep[v]--;
      f[v] = fa;
      cur--;
    }
    printf("YES\n");
    for (int i = 2; i <= n; i++) {
      printf("%d%c", f[i], i == n ? '\n' : ' ');
    }
  }
  return 0;
}

```