#include <bits/stdc++.h>
using namespace std;
int dep[5011], p[5011], L[5011], R[5011];
int N, D;
void init() {
  int i;
  for (i = 1; i <= N; ++i) {
    p[i] = L[i] = R[i] = -1;
  }
  for (i = 1; i < N; ++i) {
    p[i + 1] = i, L[i] = i + 1, dep[i] = i - 1;
  }
  dep[N] = N - 1;
}
int getv() {
  int ret = -1;
  for (int i = 2; i <= N; ++i) {
    if (L[i] == -1 && R[i] == -1) {
      if (ret == -1 || dep[ret] < dep[i]) ret = i;
    }
  }
  return ret;
}
int getnew(int limit, int up) {
  int ret = -1;
  for (int i = 1; i <= N; ++i) {
    if (dep[i] >= limit && dep[i] <= up && (L[i] == -1 || R[i] == -1)) {
      if (ret == -1 || dep[ret] > dep[i]) {
        ret = i;
      }
    }
  }
  return ret;
}
void del(int v) {
  if (p[v] != -1) {
    if (L[p[v]] == v) {
      L[p[v]] = -1;
    } else {
      R[p[v]] = -1;
    }
    p[v] = -1;
  }
}
void addchild(int v, int u) {
  if (L[v] == -1) {
    L[v] = u;
    p[u] = v;
  } else if (R[v] == -1) {
    R[v] = u;
    p[u] = v;
  }
  dep[u] = dep[v] + 1;
}
int gettol() {
  int ret = 0;
  for (int i = 1; i <= N; ++i) ret += dep[i];
  return ret;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T-- > 0) {
    scanf("%d%d", &N, &D);
    init();
    int curr = gettol();
    while (curr > D) {
      int v = getv();
      if (v == -1) break;
      int d = dep[v];
      int low = d - (curr - D) - 1, high = d - 2;
      int u = getnew(low, high);
      if (u == -1) break;
      del(v);
      addchild(u, v);
      curr = gettol();
    }
    if (curr != D) {
      printf("NO\n");
    } else {
      printf("YES\n");
      for (int i = 2; i <= N; ++i) {
        printf("%d ", p[i]);
      }
      printf("\n");
    }
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
int dep[5011], p[5011], L[5011], R[5011];
int N, D;
void init() {
  int i;
  for (i = 1; i <= N; ++i) {
    p[i] = L[i] = R[i] = -1;
  }
  for (i = 1; i < N; ++i) {
    p[i + 1] = i, L[i] = i + 1, dep[i] = i - 1;
  }
  dep[N] = N - 1;
}
int getv() {
  int ret = -1;
  for (int i = 2; i <= N; ++i) {
    if (L[i] == -1 && R[i] == -1) {
      if (ret == -1 || dep[ret] < dep[i]) ret = i;
    }
  }
  return ret;
}
int getnew(int limit, int up) {
  int ret = -1;
  for (int i = 1; i <= N; ++i) {
    if (dep[i] >= limit && dep[i] <= up && (L[i] == -1 || R[i] == -1)) {
      if (ret == -1 || dep[ret] > dep[i]) {
        ret = i;
      }
    }
  }
  return ret;
}
void del(int v) {
  if (p[v] != -1) {
    if (L[p[v]] == v) {
      L[p[v]] = -1;
    } else {
      R[p[v]] = -1;
    }
    p[v] = -1;
  }
}
void addchild(int v, int u) {
  if (L[v] == -1) {
    L[v] = u;
    p[u] = v;
  } else if (R[v] == -1) {
    R[v] = u;
    p[u] = v;
  }
  dep[u] = dep[v] + 1;
}
int gettol() {
  int ret = 0;
  for (int i = 1; i <= N; ++i) ret += dep[i];
  return ret;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T-- > 0) {
    scanf("%d%d", &N, &D);
    init();
    int curr = gettol();
    while (curr > D) {
      int v = getv();
      if (v == -1) break;
      int d = dep[v];
      int low = d - (curr - D) - 1, high = d - 2;
      int u = getnew(low, high);
      if (u == -1) break;
      del(v);
      addchild(u, v);
      curr = gettol();
    }
    if (curr != D) {
      printf("NO\n");
    } else {
      printf("YES\n");
      for (int i = 2; i <= N; ++i) {
        printf("%d ", p[i]);
      }
      printf("\n");
    }
  }
  return 0;
}

```