#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5e6 + 10, MAXM = 2e7 + 10;
const int INF = INT_MAX, SINF = 0x3f3f3f3f;
const long long llINF = LLONG_MAX;
const int MOD = 1e9 + 7, mod = 998244353;
long long f[MAXN], n, d;
long long calc(int n) {
  long long ans = 0;
  for (int i = 0; n; i++) {
    ans += i * min((1 << i), n);
    n -= min((1 << i), n);
  }
  return ans;
}
int num[5005], cur, ans[5005], cnt[5005];
vector<int> v[5005];
int solve() {
  memset(cnt, 0, sizeof(cnt));
  scanf("%lld%lld", &n, &d);
  if (d < calc(n)) return puts("NO"), 0;
  if (d > n * (n - 1) / 2) return puts("NO"), 0;
  puts("YES");
  for (int i = 1; i <= n; i++) f[i] = i - 1, num[i - 1] = 1;
  for (int i = 0; i <= n; i++) v[i].resize(0);
  int ansnow = n * (n - 1) / 2 - d;
  for (int i = n; i; i--) {
    for (int j = 0; j < i; j++) {
      if (num[j] < (1 << j) && ansnow >= i - j - 1) {
        ansnow -= (i - j - 1);
        f[i] = j;
        num[j]++;
        num[i - 1]--;
        break;
      }
    }
  }
  for (int i = 1; i <= n; i++) v[f[i]].push_back(i);
  for (int i = 1; i <= n; i++) {
    for (auto j : v[i]) {
      ans[j] = v[i - 1][cnt[i - 1] / 2];
      cnt[i - 1]++;
    }
  }
  for (int i = 2; i <= n; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
  return 0;
}

### Prompt

Please provide a cpp coded solution 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 MAXN = 5e6 + 10, MAXM = 2e7 + 10;
const int INF = INT_MAX, SINF = 0x3f3f3f3f;
const long long llINF = LLONG_MAX;
const int MOD = 1e9 + 7, mod = 998244353;
long long f[MAXN], n, d;
long long calc(int n) {
  long long ans = 0;
  for (int i = 0; n; i++) {
    ans += i * min((1 << i), n);
    n -= min((1 << i), n);
  }
  return ans;
}
int num[5005], cur, ans[5005], cnt[5005];
vector<int> v[5005];
int solve() {
  memset(cnt, 0, sizeof(cnt));
  scanf("%lld%lld", &n, &d);
  if (d < calc(n)) return puts("NO"), 0;
  if (d > n * (n - 1) / 2) return puts("NO"), 0;
  puts("YES");
  for (int i = 1; i <= n; i++) f[i] = i - 1, num[i - 1] = 1;
  for (int i = 0; i <= n; i++) v[i].resize(0);
  int ansnow = n * (n - 1) / 2 - d;
  for (int i = n; i; i--) {
    for (int j = 0; j < i; j++) {
      if (num[j] < (1 << j) && ansnow >= i - j - 1) {
        ansnow -= (i - j - 1);
        f[i] = j;
        num[j]++;
        num[i - 1]--;
        break;
      }
    }
  }
  for (int i = 1; i <= n; i++) v[f[i]].push_back(i);
  for (int i = 1; i <= n; i++) {
    for (auto j : v[i]) {
      ans[j] = v[i - 1][cnt[i - 1] / 2];
      cnt[i - 1]++;
    }
  }
  for (int i = 2; i <= n; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
  return 0;
}

```