#include <bits/stdc++.h>
void rd(int &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
void lrd(long long &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
const int INF = 1e9;
const long long LINF = 1e18;
const int N = 5e3 + 10;
using namespace std;
int T, n, d;
struct node {
  int p, dep, id, siz;
  bool operator<(const node o) const {
    return dep < o.dep || (dep == o.dep && siz > o.siz);
  }
} a[N];
bool cmp(node a, node b) { return a.id < b.id; }
int hs[N];
int main() {
  rd(T);
  while (T--) {
    rd(n);
    rd(d);
    int ans = 0;
    memset(a, 0, sizeof(a));
    a[0].dep = -1;
    for (int i = 1; i <= n; i++)
      a[i].p = i / 2, a[i].dep = a[i / 2].dep + 1, ans += a[i].dep, a[i].id = i,
      a[i / 2].siz++;
    if (ans > d || (d > n * (n - 1) / 2))
      printf("NO\n");
    else {
      while (ans < d) {
        sort(a + 1, a + n + 1);
        for (int i = 1; i <= n; i++) hs[a[i].id] = i;
        for (int i = n; i > 1; i--) {
          if (a[i].dep == a[i - 1].dep) {
            a[hs[a[i].p]].siz--;
            a[i].p = a[i - 1].id;
            a[i].dep++;
            a[i - 1].siz++;
            ans++;
            break;
          }
        }
      }
      printf("YES\n");
      sort(a + 1, a + n + 1, cmp);
      for (int i = 2; i <= n; i++) printf("%d ", a[i].p);
      puts("");
    }
  }
  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>
void rd(int &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
void lrd(long long &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
const int INF = 1e9;
const long long LINF = 1e18;
const int N = 5e3 + 10;
using namespace std;
int T, n, d;
struct node {
  int p, dep, id, siz;
  bool operator<(const node o) const {
    return dep < o.dep || (dep == o.dep && siz > o.siz);
  }
} a[N];
bool cmp(node a, node b) { return a.id < b.id; }
int hs[N];
int main() {
  rd(T);
  while (T--) {
    rd(n);
    rd(d);
    int ans = 0;
    memset(a, 0, sizeof(a));
    a[0].dep = -1;
    for (int i = 1; i <= n; i++)
      a[i].p = i / 2, a[i].dep = a[i / 2].dep + 1, ans += a[i].dep, a[i].id = i,
      a[i / 2].siz++;
    if (ans > d || (d > n * (n - 1) / 2))
      printf("NO\n");
    else {
      while (ans < d) {
        sort(a + 1, a + n + 1);
        for (int i = 1; i <= n; i++) hs[a[i].id] = i;
        for (int i = n; i > 1; i--) {
          if (a[i].dep == a[i - 1].dep) {
            a[hs[a[i].p]].siz--;
            a[i].p = a[i - 1].id;
            a[i].dep++;
            a[i - 1].siz++;
            ans++;
            break;
          }
        }
      }
      printf("YES\n");
      sort(a + 1, a + n + 1, cmp);
      for (int i = 2; i <= n; i++) printf("%d ", a[i].p);
      puts("");
    }
  }
  return 0;
}

```