#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class T>
using V = vector<T>;
template <class T, class U>
using P = pair<T, U>;
using vll = V<ll>;
using vvll = V<vll>;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
constexpr ll MOD = 1000000007;
constexpr ll HIGHINF = (ll)1e18;
constexpr int INF = 1e9;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t-- > 0) {
    int n, d;
    cin >> n >> d;
    int mind = 0, tmpd = 0;
    for (int i = 2; i <= n; i++) {
      if (__builtin_popcount(i) == 1) tmpd++;
      mind += tmpd;
    }
    if (mind > d || d > n * (n - 1) / 2) {
      cout << "NO\n";
      continue;
    }
    V<int> par(n, -1), chc(n, 1), depth(n, 0);
    for (int i = 1; i < n; i++) {
      par[i] = i - 1;
      depth[i] = i;
    }
    chc[n - 1] = 0;
    V<bool> move(n, true);
    int tmp = n * (n - 1) / 2;
    while (tmp > d) {
      int leaf = -1;
      for (int i = 0; i < n; i++) {
        if (move[i] && chc[i] == 0 && (leaf == -1 || depth[leaf] > depth[i]))
          leaf = i;
      }
      int lp = -1;
      for (int i = 0; i < n; i++) {
        if (chc[i] <= 1 && depth[i] == depth[leaf] - 2 &&
            (lp == -1 || depth[lp] < depth[i]))
          lp = i;
      }
      if (lp == -1) {
        move[leaf] = false;
        continue;
      }
      chc[par[leaf]]--;
      chc[lp]++;
      depth[leaf]--;
      par[leaf] = lp;
      tmp--;
    }
    cout << "YES\n";
    for (int i = 1; i < n; i++) cout << par[i] + 1 << (i == n - 1 ? '\n' : ' ');
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following 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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class T>
using V = vector<T>;
template <class T, class U>
using P = pair<T, U>;
using vll = V<ll>;
using vvll = V<vll>;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
constexpr ll MOD = 1000000007;
constexpr ll HIGHINF = (ll)1e18;
constexpr int INF = 1e9;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t-- > 0) {
    int n, d;
    cin >> n >> d;
    int mind = 0, tmpd = 0;
    for (int i = 2; i <= n; i++) {
      if (__builtin_popcount(i) == 1) tmpd++;
      mind += tmpd;
    }
    if (mind > d || d > n * (n - 1) / 2) {
      cout << "NO\n";
      continue;
    }
    V<int> par(n, -1), chc(n, 1), depth(n, 0);
    for (int i = 1; i < n; i++) {
      par[i] = i - 1;
      depth[i] = i;
    }
    chc[n - 1] = 0;
    V<bool> move(n, true);
    int tmp = n * (n - 1) / 2;
    while (tmp > d) {
      int leaf = -1;
      for (int i = 0; i < n; i++) {
        if (move[i] && chc[i] == 0 && (leaf == -1 || depth[leaf] > depth[i]))
          leaf = i;
      }
      int lp = -1;
      for (int i = 0; i < n; i++) {
        if (chc[i] <= 1 && depth[i] == depth[leaf] - 2 &&
            (lp == -1 || depth[lp] < depth[i]))
          lp = i;
      }
      if (lp == -1) {
        move[leaf] = false;
        continue;
      }
      chc[par[leaf]]--;
      chc[lp]++;
      depth[leaf]--;
      par[leaf] = lp;
      tmp--;
    }
    cout << "YES\n";
    for (int i = 1; i < n; i++) cout << par[i] + 1 << (i == n - 1 ? '\n' : ' ');
  }
  return 0;
}

```