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>
template <class C, class E>
inline bool contains(const C& container, const E& element) {
  return container.find(element) != container.end();
}
template <class T>
inline void checkmin(T& a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T& a, T b) {
  if (b > a) a = b;
}
using namespace std;
int Sum(int a) { return (a * (a + 1)) / 2; }
vector<int> degSum;
int LevelCount(int l) { return degSum[l - 1]; }
bool GotLeave(int l) {
  return LevelCount(l) > 0 && LevelCount(l + 1) <= LevelCount(l) * 2 - 2;
}
bool GotSpace(int l) {
  return LevelCount(l) > 0 && LevelCount(l + 1) < LevelCount(l) * 2 - 2;
}
int n, d;
void Inc() {
  for (int l = 1; l < degSum.size(); ++l) {
    if (degSum[l - 1] == 0) {
      assert(false);
      return;
    }
    if (LevelCount(l) > 1 && GotLeave(l) && GotSpace(l)) {
      degSum[l - 1]--;
      degSum[l]++;
      return;
    }
  }
}
void Trace() {
  cout << "YES" << endl;
  int rem = n - 1;
  int level = 0;
  int firstNodeLastLEvel = 1;
  int nextNode = 2;
  while (degSum[level] != 0) {
    int firstNodeThisLevel = nextNode;
    int tl = min(rem, degSum[level]);
    rem -= tl;
    nextNode += tl;
    for (int i = (0), _b = ((tl)-1); i <= _b; i++)
      cout << firstNodeLastLEvel + i / 2 << " ";
    firstNodeLastLEvel = firstNodeThisLevel;
    level++;
  }
  cout << endl;
}
void Solve() {
  cin >> n >> d;
  int maxV = Sum(n - 1);
  if (d > maxV) {
    cout << "NO" << endl;
    ;
    return;
  }
  int available = 2;
  int level = 0;
  int rem = n - 1;
  int sum = 0;
  degSum.assign(200, 0);
  while (rem > 0) {
    degSum[level] = min(rem, available);
    sum += (level + 1) * degSum[level];
    rem -= degSum[level];
    level++;
    available *= 2;
  }
  if (sum > d) {
    cout << "NO" << endl;
    ;
    return;
  }
  while (sum < d) {
    Inc();
    sum++;
  }
  Trace();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << std::setprecision(15);
  cout << std::fixed;
  int t;
  cin >> t;
  for (int i = (0), _b = ((t)-1); i <= _b; i++) Solve();
  return 0;
}