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;
const int MAXN = 5005;
int valid(int n, int d) {
  int sumDepth = 0;
  int cnt = 1;
  int depth = 0;
  while (n > 0) {
    if (n < cnt) {
      sumDepth == n* depth;
      n = 0;
      break;
    }
    sumDepth += cnt * depth;
    n -= cnt;
    cnt *= 2;
    depth++;
  }
  return sumDepth <= d && d <= n * (n - 1) / 2;
}
int adj[MAXN][2];
int depth[MAXN];
int par[MAXN];
int leftFree[MAXN];
int rightFree[MAXN];
void solve() {
  memset(adj, 0, sizeof(adj));
  memset(depth, 0, sizeof(depth));
  memset(par, 0, sizeof(par));
  int n, d;
  cin >> n >> d;
  depth[0] = 0;
  for (int i = 2; i <= n; i++) {
    adj[i / 2][i % 2] = i;
    par[i] = i / 2;
    depth[i] = depth[i / 2] + 1;
  }
  for (int i = 2; i <= n; i++) {
    d -= depth[i];
  }
  if (d < 0) {
    cout << "NO\n";
    return;
  }
  while (d--) {
    memset(leftFree, 0, sizeof(leftFree));
    memset(rightFree, 0, sizeof(rightFree));
    for (int i = 1; i <= n; i++) {
      if (adj[i][0] == 0 && adj[i][1] == 0) {
        int p = par[i];
        if (i == adj[p][0]) {
          leftFree[depth[i]] = i;
        } else {
          rightFree[depth[i]] = i;
        }
      }
    }
    int done = 0;
    for (int i = 1; i <= n; i++) {
      if (adj[i][0] == 0) {
        if (leftFree[depth[i]] != 0 && i != leftFree[depth[i]]) {
          int oldP = par[leftFree[depth[i]]];
          adj[oldP][0] = 0;
          adj[i][0] = leftFree[depth[i]];
          depth[adj[i][0]] = depth[i] + 1;
          par[adj[i][0]] = i;
          done = 1;
        } else if (rightFree[depth[i]] != 0 && i != rightFree[depth[i]]) {
          int oldP = par[rightFree[depth[i]]];
          adj[oldP][1] = 0;
          adj[i][0] = rightFree[depth[i]];
          depth[adj[i][0]] = depth[i] + 1;
          par[adj[i][0]] = i;
          done = 1;
        }
      } else if (adj[i][1] == 0) {
        if (leftFree[depth[i]] != 0 && i != leftFree[depth[i]]) {
          int oldP = par[leftFree[depth[i]]];
          adj[oldP][0] = 0;
          adj[i][1] = leftFree[depth[i]];
          depth[adj[i][1]] = depth[i] + 1;
          par[adj[i][1]] = i;
          done = 1;
        } else if (rightFree[depth[i]] != 0 && i != rightFree[depth[i]]) {
          int oldP = par[rightFree[depth[i]]];
          adj[oldP][1] = 0;
          adj[i][1] = rightFree[depth[i]];
          depth[adj[i][1]] = depth[i] + 1;
          par[adj[i][1]] = i;
          done = 1;
        }
      }
      if (done) {
        break;
      }
    }
    if (!done) {
      cout << "NO\n";
      return;
    }
  }
  cout << "YES\n";
  for (int i = 2; i <= n; i++) {
    cout << par[i] << " \n"[i == n];
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
}
