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>
inline void read(_t &x) {
  x = 0;
  _t fu = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') fu = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch & 15);
    ch = getchar();
  }
  x *= fu;
}
long long gcd(long long x, long long y) { return !x ? y : gcd(y % x, x); }
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, a % b, y, x);
  y -= a / b * x;
}
struct Mart {
  int a, b, c, d;
  Mart operator*(Mart &m) {
    Mart temp;
    temp.a = a * m.a + b * m.c;
    temp.b = a * m.b + b * m.d;
    temp.c = c * m.a + d * m.c;
    temp.d = c * m.b + d * m.d;
    return temp;
  }
  Mart() { a = b = c = d = 0; }
  Mart(int x[4]) {
    a = x[0];
    b = x[1];
    c = x[2];
    d = x[3];
  }
  Mart(int u, int v, int w, int x) {
    a = u;
    b = v;
    c = w;
    d = x;
  }
  int det() { return a * d - b * c; }
};
const int maxn = 5e3 + 10;
int n, d;
int cnt[maxn];
void de() {
  int sum = 0;
  for (int i = 0; i < n; i++) {
    printf("%d ", cnt[i]);
    sum += cnt[i] * i;
  }
  printf(": total %d", sum);
  printf("\n");
}
vector<int> vec, temp;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    vec.clear();
    temp.clear();
    scanf("%d%d", &n, &d);
    if (d > n * (n - 1) / 2) {
      puts("NO");
      continue;
    } else if (d == n * (n - 1) / 2) {
      puts("YES");
      for (int i = 0; i < n - 1; i++) printf("%d ", i + 1);
      printf("\n");
      continue;
    }
    int left = n * (n - 1) / 2 - d;
    for (int i = 0; i < n; i++) cnt[i] = 1;
    for (int l = 1, r = n - 1; l < r; r--) {
      while (l < r && (cnt[l] == cnt[l - 1] * 2 || left - r + l < 0)) l++;
      if (l >= r) break;
      left += l - r;
      cnt[l]++;
      cnt[r]--;
    }
    if (left)
      puts("NO");
    else {
      puts("YES");
      int id = 2, now = 1;
      vec.push_back(1);
      vec.push_back(1);
      while (now < n) {
        for (int i = 0; i < cnt[now]; i++) {
          printf("%d ", vec[i]);
          temp.push_back(id);
          temp.push_back(id);
          id++;
        }
        vec = temp;
        temp.clear();
        now++;
      }
      printf("\n");
    }
  }
  return 0;
}