#include <bits/stdc++.h>
int n, d, sum, node, num[5001], now, son[5001], fa[5001], cnt, lst, fi[5001];
inline int min(int a, int b) { return a < b ? a : b; }
inline void solve() {
  memset(son, 0, sizeof(son));
  memset(fa, 0, sizeof(fa));
  memset(fi, 0x3f, sizeof(fi));
  scanf("%d %d", &n, &d);
  node = 0;
  sum = 0;
  for (int i = 0; i <= 31; i++) {
    if ((n - node) <= (1 << i)) {
      sum += (n - node) * i;
      break;
    }
    node += (1 << i);
    sum += (i * (1 << i));
  }
  if (sum > d) {
    puts("NO");
    return;
  }
  if (d > ((n * (n - 1)) >> 1)) {
    puts("NO");
    return;
  }
  sum = 0;
  for (int i = 0; i <= n - 1; i++) num[i] = 1, sum += i;
  now = 1;
  for (int i = n - 1; i >= 0; i--) {
    if ((sum - (i - now)) <= d) {
      num[i - (sum - d)]++;
      num[i]--;
      break;
    }
    num[i]--;
    num[now]++;
    sum -= (i - now);
    if (num[now] == (1 << now)) now++;
  }
  cnt = 1;
  lst = 1;
  for (int i = 1; i <= n - 1; i++) {
    if (!num[i]) break;
    for (int j = 1; j <= num[i]; j++) {
      son[lst]++;
      fa[++cnt] = lst;
      fi[i] = min(fi[i], cnt);
      if (son[lst] == 2) lst++;
    }
    lst = fi[i];
  }
  puts("YES");
  for (int i = 2; i <= n; i++) printf("%d ", fa[i]);
  puts("");
}
int main() {
  int TEST;
  scanf("%d", &TEST);
  while (TEST--) solve();
}

### Prompt

Create a solution in CPP for the following 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>

### Response

```cpp
#include <bits/stdc++.h>
int n, d, sum, node, num[5001], now, son[5001], fa[5001], cnt, lst, fi[5001];
inline int min(int a, int b) { return a < b ? a : b; }
inline void solve() {
  memset(son, 0, sizeof(son));
  memset(fa, 0, sizeof(fa));
  memset(fi, 0x3f, sizeof(fi));
  scanf("%d %d", &n, &d);
  node = 0;
  sum = 0;
  for (int i = 0; i <= 31; i++) {
    if ((n - node) <= (1 << i)) {
      sum += (n - node) * i;
      break;
    }
    node += (1 << i);
    sum += (i * (1 << i));
  }
  if (sum > d) {
    puts("NO");
    return;
  }
  if (d > ((n * (n - 1)) >> 1)) {
    puts("NO");
    return;
  }
  sum = 0;
  for (int i = 0; i <= n - 1; i++) num[i] = 1, sum += i;
  now = 1;
  for (int i = n - 1; i >= 0; i--) {
    if ((sum - (i - now)) <= d) {
      num[i - (sum - d)]++;
      num[i]--;
      break;
    }
    num[i]--;
    num[now]++;
    sum -= (i - now);
    if (num[now] == (1 << now)) now++;
  }
  cnt = 1;
  lst = 1;
  for (int i = 1; i <= n - 1; i++) {
    if (!num[i]) break;
    for (int j = 1; j <= num[i]; j++) {
      son[lst]++;
      fa[++cnt] = lst;
      fi[i] = min(fi[i], cnt);
      if (son[lst] == 2) lst++;
    }
    lst = fi[i];
  }
  puts("YES");
  for (int i = 2; i <= n; i++) printf("%d ", fa[i]);
  puts("");
}
int main() {
  int TEST;
  scanf("%d", &TEST);
  while (TEST--) solve();
}

```