Codeforces 1311 E. Construct the Binary Tree（构造二叉树）

Description：

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:

解题思路：

给出 $n$ 个节点深度和为 $d$ 问你是否可以构成一个二叉树，可以的话输出全部节点深度。
全部节点组成一条链的话深度和肯定最大，我们就先组成一条链，然后每次移动最下面那个节点，看看移动到哪里合适，然后判断是否可以组成深度和为 $d$。

AC代码：

const int N = 5e3 + 5;
int n, d;
int ans[N];
int ch[N];
int fa[N];
int c, now, mx;
int cnt, pos;
int get_dep()
{
    int ans = now;
    c++;
    if (c == mx)
        now++, mx *= 2, c = 1;
    return ans;
}

int main()
{
    int t;
    sd(t);
    while (t--)
    {
        mem(ch, 0);
        sdd(n, d);
        cnt = 0;
        rep(i, 1, n - 1)
        {
            ans[i] = i;
            cnt += i;
        }
        pos = n - 1;
        now = 1, c = 1, mx = 2;
        if (cnt < d)
        {
            puts("NO");
            continue;
        }
        while (cnt > d)
        {
            int dep = get_dep();
            if (dep >= pos)
                break;
            int tmp = cnt - d;
            if (tmp >= (pos - dep))
            {
                cnt -= pos - dep;
                ans[pos] = dep;
            }
            else
            {
                ans[pos] -= tmp;
                cnt = d;
            }
            pos--;
        }
        if (cnt > d)
            puts("NO");
        else
        {
            puts("YES");
            ans[0] = 0;
            rep(i, 1, n - 1)
            {
                rep(j, 0, n - 1) if (ans[j] == ans[i] - 1 && ch[j] < 2)
                {
                    ch[j]++;
                    fa[i] = j;
                    break;
                }
            }
            rep(i, 1, n - 1)
            {
                printf("%d%c", fa[i] + 1, i == n - 1 ? '\n' : ' ');
            }
        }
    }
    return 0;
}