#include <bits/stdc++.h>
int main() {
  unsigned long t;
  std::cin >> t;
  std::vector<unsigned long> countForLevel;
  for (; t > 0; t--) {
    unsigned long n;
    long d;
    std::cin >> n >> d;
    countForLevel.clear();
    countForLevel.push_back(1);
    n--;
    while (n > 0) {
      unsigned long newDepth = countForLevel.size();
      unsigned long max = 2 * countForLevel.at(newDepth - 1);
      unsigned long delta = (n <= max ? n : max);
      countForLevel.push_back(delta);
      d -= newDepth * delta;
      n -= delta;
    }
    if (d < 0) {
      std::cout << "NO\n";
    } else {
      unsigned long ud = (unsigned long)d;
      for (unsigned long row = countForLevel.size() - 1; ud > 0 && row > 0;
           row--) {
        while (countForLevel.at(row) > 1 && ud > 0) {
          unsigned long M = countForLevel.size();
          countForLevel.at(row)--;
          if (ud + row >= M) {
            countForLevel.push_back(1);
            ud -= (M - row);
          } else {
            countForLevel.at(row + ud)++;
            ud = 0;
          }
        }
      }
      if (ud > 0) {
        std::cout << "NO\n";
      } else {
        std::cout << "YES\n";
        long currParent = 1;
        long firstChild = 2;
        std::string prefix = "";
        for (unsigned long i = 1; i < countForLevel.size(); i++) {
          unsigned long N = countForLevel.at(i);
          long childCount = 0;
          long nextChild = firstChild;
          for (unsigned long z = 0; z < N; z++) {
            std::cout << prefix << currParent;
            prefix = " ";
            nextChild++;
            childCount++;
            if (childCount == 2) {
              childCount = 0;
              currParent++;
            }
          }
          currParent = firstChild;
          firstChild = nextChild;
        }
        std::cout << "\n";
      }
    }
  }
}

### 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>
int main() {
  unsigned long t;
  std::cin >> t;
  std::vector<unsigned long> countForLevel;
  for (; t > 0; t--) {
    unsigned long n;
    long d;
    std::cin >> n >> d;
    countForLevel.clear();
    countForLevel.push_back(1);
    n--;
    while (n > 0) {
      unsigned long newDepth = countForLevel.size();
      unsigned long max = 2 * countForLevel.at(newDepth - 1);
      unsigned long delta = (n <= max ? n : max);
      countForLevel.push_back(delta);
      d -= newDepth * delta;
      n -= delta;
    }
    if (d < 0) {
      std::cout << "NO\n";
    } else {
      unsigned long ud = (unsigned long)d;
      for (unsigned long row = countForLevel.size() - 1; ud > 0 && row > 0;
           row--) {
        while (countForLevel.at(row) > 1 && ud > 0) {
          unsigned long M = countForLevel.size();
          countForLevel.at(row)--;
          if (ud + row >= M) {
            countForLevel.push_back(1);
            ud -= (M - row);
          } else {
            countForLevel.at(row + ud)++;
            ud = 0;
          }
        }
      }
      if (ud > 0) {
        std::cout << "NO\n";
      } else {
        std::cout << "YES\n";
        long currParent = 1;
        long firstChild = 2;
        std::string prefix = "";
        for (unsigned long i = 1; i < countForLevel.size(); i++) {
          unsigned long N = countForLevel.at(i);
          long childCount = 0;
          long nextChild = firstChild;
          for (unsigned long z = 0; z < N; z++) {
            std::cout << prefix << currParent;
            prefix = " ";
            nextChild++;
            childCount++;
            if (childCount == 2) {
              childCount = 0;
              currParent++;
            }
          }
          currParent = firstChild;
          firstChild = nextChild;
        }
        std::cout << "\n";
      }
    }
  }
}

```