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:

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#include <bits/stdc++.h>
using namespace std;
int parents[5000];
int leftNeighbor[5000];
int children[5000][2];
void initParents() {
  for (int i = 0; i < 5000; i++) {
    parents[i] = i - 1;
  }
}
void initChildren() {
  for (int i = 0; i < 5000; i++) {
    children[i][0] = i + 1;
    children[i][1] = -1;
  }
}
void initLeftNeighbor() {
  for (int i = 0; i < 5000; i++) {
    leftNeighbor[i] = -1;
  }
}
int getMin(int current, int depthP2, int depthP1) {
  return current - (depthP2) + depthP1 + 1;
}
int main() {
  int T;
  cin >> T;
  for (int t = 0; t < T; t++) {
    int n, d;
    cin >> n >> d;
    int current = ((n - 1) * n) / 2;
    if (d > current) {
      cout << "NO\n";
      continue;
    }
    initParents();
    initChildren();
    children[n - 1][0] = -1;
    children[n - 1][1] = -1;
    initLeftNeighbor();
    int pointer1 = 0;
    int pointer2 = n - 1;
    int dp1 = 0;
    int dp2 = n - 1;
    int firstInRow = 0;
    int previous = -1;
    bool possible = false;
    while (current != d) {
      int minPossible = getMin(current, dp2, dp1);
      if (minPossible >= current) {
        break;
      }
      if (minPossible >= d) {
        int oldParent = parents[pointer2];
        children[parents[pointer2]][0] = -1;
        parents[pointer2] = pointer1;
        int child = children[pointer1][1] == -1 ? 1 : 0;
        children[pointer1][child] = pointer2;
        if (previous != -1) {
          leftNeighbor[previous] = pointer2;
        }
        bool updateP1 =
            children[pointer1][0] != -1 && children[pointer1][1] != -1;
        bool startNewRow = child == 1 && children[pointer1][0] != -1;
        previous = pointer2;
        if (updateP1) {
          if (startNewRow) {
            leftNeighbor[pointer2] = children[pointer1][0];
            pointer1 = children[firstInRow][1];
            firstInRow = pointer1;
            previous = -1;
            dp1++;
          } else {
            pointer1 = leftNeighbor[pointer1];
          }
        }
        current = minPossible;
        pointer2 = oldParent;
        dp2--;
      } else {
        int need = current - d;
        int newParent = parents[pointer2];
        for (int i = 0; i < need; i++) {
          newParent = parents[newParent];
        }
        parents[pointer2] = newParent;
        current = d;
      }
    }
    if (current == d) {
      cout << "YES\n";
      for (int i = 1; i < n; i++) {
        cout << (parents[i] + 1) << " ";
      }
      cout << '\n';
    } else {
      cout << "NO\n";
    }
  }
}
