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) {
  x = 0;
  char ch = getchar();
  long long f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= f;
}
template <typename T, typename... Args>
void read(T &first, Args &...args) {
  read(first);
  read(args...);
}
template <typename T>
void write(T arg) {
  T x = arg;
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
template <typename T, typename... Ts>
void write(T arg, Ts... args) {
  write(arg);
  if (sizeof...(args) != 0) {
    putchar(' ');
    write(args...);
  }
}
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const int N = 1e5 + 5;
vector<int> ve;
int T, n, m, L, R;
int fa[N], dep[N], vis[N], vertex[N];
void getL() {
  long long t = 2, dis = 1, sum = n - 1;
  while (sum - t >= 0) {
    L += t * dis;
    sum -= t;
    t *= 2;
    dis += 1;
  }
  for (int i = 1; i <= sum; i++) ve.push_back(dis);
  L += sum * dis;
}
void solve() {
  int x = m, mx = 0;
  for (int i = 2; i <= n; i++)
    fa[i] = i / 2, dep[i] = dep[fa[i]] + 1, x -= dep[i], mx = max(mx, dep[i]);
  int p = n;
  while (p) {
    vertex[dep[p]] = p;
    vis[p] = 1;
    p = fa[p];
  }
  for (int i = n; i >= 1; i--) {
    if (vis[i]) continue;
    int pre = mx;
    while (dep[fa[i]] < pre && x) {
      fa[i] = vertex[dep[fa[i]] + 1];
      dep[i] = dep[fa[i]] + 1;
      if (dep[i] > mx) mx++, vertex[mx] = i, vis[i] = 1;
      x--;
    }
  }
  for (int i = 2; i <= n; i++) cout << fa[i] << ' ';
  cout << endl;
}
int main() {
  read(T);
  while (T--) {
    ve.clear();
    L = 0, R = 0;
    read(n, m);
    for (int i = 0; i <= n; i++) vertex[i] = vis[i] = dep[i] = 0;
    vis[0] = 1;
    getL();
    long long R = 1ll * n * (n - 1) / 2;
    if (L <= m && m <= R) {
      cout << "YES" << endl;
      solve();
    } else
      cout << "NO" << endl;
  }
}