#include <bits/stdc++.h>
using namespace std;
const int oo = 1e9 + 5;
const int MOD = 998244353;
const long long ooll = 3e18;
const int N = 5e3 + 5;
int mn[N], p[N];
void pre() {
  mn[0] = 0;
  for (int i = 1; i < N; ++i) {
    int dep = 0, j = i;
    while (i > 1) {
      i >>= 1;
      dep++;
    }
    i = j;
    mn[i] = mn[i - 1] + dep;
  }
}
void construct(int root, int sz, int d) {
  assert(d >= mn[sz] && d <= (sz * (sz - 1)) / 2);
  if (sz <= 1) return;
  int lef;
  for (lef = 0; lef < sz; lef++) {
    int l = lef, r = sz - 1 - lef;
    int xl = mn[l], xr = mn[r], yl = (l * (l - 1)) / 2, yr = (r * (r - 1)) / 2;
    int need = d - sz + 1;
    if (need >= xl + xr && need <= yl + yr) {
      int dl = max(need - yr, xl);
      int dr = need - dl;
      assert(dl >= xl && dl <= yl);
      assert(dr >= xr && dr <= yr);
      construct(root + 1, l, dl);
      if (l > 0) p[root + 1] = root;
      construct(root + 1 + l, r, dr);
      if (r > 0) p[root + 1 + l] = root;
      break;
    }
  }
  assert(lef < sz);
}
void solve() {
  int n, d;
  cin >> n >> d;
  if (d >= mn[n] && d <= (n * (n - 1)) / 2) {
    cout << "YES\n";
    construct(1, n, d);
    for (int i = 2; i <= n; ++i) cout << p[i] << " \n"[i == n];
  } else {
    cout << "NO\n";
  }
}
int main() {
  pre();
  int t = 1;
  cin >> t;
  for (int tc = 1; tc <= t; ++tc) {
    solve();
  }
  return 0;
}

### 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>
using namespace std;
const int oo = 1e9 + 5;
const int MOD = 998244353;
const long long ooll = 3e18;
const int N = 5e3 + 5;
int mn[N], p[N];
void pre() {
  mn[0] = 0;
  for (int i = 1; i < N; ++i) {
    int dep = 0, j = i;
    while (i > 1) {
      i >>= 1;
      dep++;
    }
    i = j;
    mn[i] = mn[i - 1] + dep;
  }
}
void construct(int root, int sz, int d) {
  assert(d >= mn[sz] && d <= (sz * (sz - 1)) / 2);
  if (sz <= 1) return;
  int lef;
  for (lef = 0; lef < sz; lef++) {
    int l = lef, r = sz - 1 - lef;
    int xl = mn[l], xr = mn[r], yl = (l * (l - 1)) / 2, yr = (r * (r - 1)) / 2;
    int need = d - sz + 1;
    if (need >= xl + xr && need <= yl + yr) {
      int dl = max(need - yr, xl);
      int dr = need - dl;
      assert(dl >= xl && dl <= yl);
      assert(dr >= xr && dr <= yr);
      construct(root + 1, l, dl);
      if (l > 0) p[root + 1] = root;
      construct(root + 1 + l, r, dr);
      if (r > 0) p[root + 1 + l] = root;
      break;
    }
  }
  assert(lef < sz);
}
void solve() {
  int n, d;
  cin >> n >> d;
  if (d >= mn[n] && d <= (n * (n - 1)) / 2) {
    cout << "YES\n";
    construct(1, n, d);
    for (int i = 2; i <= n; ++i) cout << p[i] << " \n"[i == n];
  } else {
    cout << "NO\n";
  }
}
int main() {
  pre();
  int t = 1;
  cin >> t;
  for (int tc = 1; tc <= t; ++tc) {
    solve();
  }
  return 0;
}

```