#include <bits/stdc++.h>
long long t, n, d, fa[1000005], dep[1000005];
long long read() {
  char c = getchar();
  long long ans = 0;
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') ans = ans * 10 + c - '0', c = getchar();
  return ans;
}
void Write(long long x) {
  if (x < 10)
    putchar(x ^ 48);
  else
    Write(x / 10), putchar((x % 10) ^ 48);
  return;
}
void solve() {
  long long sum = 0, pos;
  dep[0] = -1;
  for (long long i = 1; i <= n; i++) {
    fa[i] = i / 2, dep[i] = dep[i / 2] + 1, sum += dep[i];
    if (!(i & (i - 1))) pos = i;
  }
  if (d < sum) return printf("NO\n"), void();
  d -= sum;
  for (long long i = n; i && d; i--) {
    if (!(i & (i - 1))) continue;
    d -= dep[pos] + 1 - dep[i];
    if (d <= 0) {
      while (d) pos = fa[pos], ++d;
      fa[i] = pos;
    }
    fa[i] = pos, dep[i] = dep[pos] + 1, pos = i;
  }
  if (d) return printf("NO\n"), void();
  printf("YES\n");
  for (long long i = 2; i <= n; i++) Write(fa[i]), putchar(' ');
  return putchar('\n'), void();
}
int main() {
  t = read();
  while (t--) n = read(), d = read(), solve();
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
long long t, n, d, fa[1000005], dep[1000005];
long long read() {
  char c = getchar();
  long long ans = 0;
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') ans = ans * 10 + c - '0', c = getchar();
  return ans;
}
void Write(long long x) {
  if (x < 10)
    putchar(x ^ 48);
  else
    Write(x / 10), putchar((x % 10) ^ 48);
  return;
}
void solve() {
  long long sum = 0, pos;
  dep[0] = -1;
  for (long long i = 1; i <= n; i++) {
    fa[i] = i / 2, dep[i] = dep[i / 2] + 1, sum += dep[i];
    if (!(i & (i - 1))) pos = i;
  }
  if (d < sum) return printf("NO\n"), void();
  d -= sum;
  for (long long i = n; i && d; i--) {
    if (!(i & (i - 1))) continue;
    d -= dep[pos] + 1 - dep[i];
    if (d <= 0) {
      while (d) pos = fa[pos], ++d;
      fa[i] = pos;
    }
    fa[i] = pos, dep[i] = dep[pos] + 1, pos = i;
  }
  if (d) return printf("NO\n"), void();
  printf("YES\n");
  for (long long i = 2; i <= n; i++) Write(fa[i]), putchar(' ');
  return putchar('\n'), void();
}
int main() {
  t = read();
  while (t--) n = read(), d = read(), solve();
  return 0;
}

```