#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ '0'), c = getchar();
  return x * f;
}
const int N = 1e5 + 5;
int n, d, f[N], dep[N];
inline void solve() {
  n = read(), d = read();
  if (n * (n - 1) / 2 < d) return puts("NO"), void();
  queue<int> q;
  q.push(1);
  int j = 1;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    if (j < n) f[++j] = x, dep[j] = dep[x] + 1, q.push(j);
    if (j < n) f[++j] = x, dep[j] = dep[x] + 1, q.push(j);
    if (j == n) break;
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) ans += dep[i];
  if (ans > d) return puts("NO"), void();
  d -= ans;
  if (!d) {
    puts("YES");
    for (int i = 2; i <= n; i++) printf("%d ", f[i]);
    puts("");
    return;
  }
  int k = 1;
  while (k << 1 <= n) k <<= 1;
  for (int i = n; i >= 1; i--) {
    if (!(i & (i - 1))) continue;
    if (dep[k] + 1 - dep[i] < d)
      d -= dep[k] + 1 - dep[i], f[i] = k, dep[i] = dep[k] + 1, k = i;
    else {
      int x = k;
      while (dep[x] + 1 - dep[i] > d) x = f[x];
      d -= dep[x] + 1 - dep[i];
      f[i] = x;
      dep[i] = dep[x] + 1;
      k = i;
      break;
    }
    if (!d) break;
  }
  if (d)
    puts("NO");
  else {
    puts("YES");
    for (int i = 2; i <= n; i++) printf("%d ", f[i]);
    puts("");
  }
}
int main() {
  int T = read();
  while (T--) solve();
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to 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>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ '0'), c = getchar();
  return x * f;
}
const int N = 1e5 + 5;
int n, d, f[N], dep[N];
inline void solve() {
  n = read(), d = read();
  if (n * (n - 1) / 2 < d) return puts("NO"), void();
  queue<int> q;
  q.push(1);
  int j = 1;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    if (j < n) f[++j] = x, dep[j] = dep[x] + 1, q.push(j);
    if (j < n) f[++j] = x, dep[j] = dep[x] + 1, q.push(j);
    if (j == n) break;
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) ans += dep[i];
  if (ans > d) return puts("NO"), void();
  d -= ans;
  if (!d) {
    puts("YES");
    for (int i = 2; i <= n; i++) printf("%d ", f[i]);
    puts("");
    return;
  }
  int k = 1;
  while (k << 1 <= n) k <<= 1;
  for (int i = n; i >= 1; i--) {
    if (!(i & (i - 1))) continue;
    if (dep[k] + 1 - dep[i] < d)
      d -= dep[k] + 1 - dep[i], f[i] = k, dep[i] = dep[k] + 1, k = i;
    else {
      int x = k;
      while (dep[x] + 1 - dep[i] > d) x = f[x];
      d -= dep[x] + 1 - dep[i];
      f[i] = x;
      dep[i] = dep[x] + 1;
      k = i;
      break;
    }
    if (!d) break;
  }
  if (d)
    puts("NO");
  else {
    puts("YES");
    for (int i = 2; i <= n; i++) printf("%d ", f[i]);
    puts("");
  }
}
int main() {
  int T = read();
  while (T--) solve();
  return 0;
}

```