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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x);
template <typename T>
void write(T x);
template <typename T>
void writesp(T x);
template <typename T>
void writeln(T x);
const long long N = 5500;
long long q;
long long n, D;
long long h[N];
long long sum;
long long cnt;
long long tmp;
inline void clear() {
  memset(h, 0, sizeof h);
  h[cnt = 1] = 1;
  sum = 0;
}
inline void solve() {
  read(n);
  read(D);
  clear();
  for (register long long i = 2; i <= n; i++) {
    if (h[cnt] >= h[cnt - 1] * 2) ++cnt;
    ++h[cnt];
    sum += cnt - 1;
  }
  if (D < sum) return (void)(puts("NO"));
  while (sum < D) {
    tmp = 0;
    for (register long long i = 1; i <= cnt; i++) {
      if ((h[i] - 1) * 2 > h[i + 1]) {
        tmp = i;
        break;
      }
    }
    if (!tmp) break;
    if (tmp == cnt) ++cnt;
    ++sum;
    --h[tmp];
    ++h[tmp + 1];
  }
  if (sum != D) return (void)(puts("NO"));
  puts("YES");
  sum = 0;
  for (register long long i = 2; i <= cnt; i++) {
    sum += h[i - 2];
    for (register long long j = 1, tot = 0; j <= h[i]; j++) {
      if (j & 1) ++tot;
      writesp(sum + tot);
    }
  }
  putchar('\n');
}
int main() {
  read(q);
  while (q--) solve();
}
template <typename T>
void read(T &x) {
  char wn = getchar();
  long long t = 1;
  x = 0;
  while (wn < '0' || wn > '9') {
    if (wn == '-') t = -1;
    wn = getchar();
  }
  while (wn >= '0' && wn <= '9') {
    x = x * 10 + wn - '0';
    wn = getchar();
  }
  x *= t;
}
template <typename T>
void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x <= 9) {
    putchar(x + '0');
    return;
  }
  write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  putchar('\n');
}
template <typename T>
void writesp(T x) {
  write(x);
  putchar(' ');
}