Construct the Binary Tree

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 p2,p3,…,pn in the second line, where pi is the parent of the vertex i. Note that the sequence of parents you print should describe some binary tree.

Example

inputCopy
3
5 7
10 19
10 18
outputCopy
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:

#include <bits/stdc++.h>

using namespace std;
const int N = 5e3 + 10;
int cnt[N], fa[N], t, n, d;
vector<int> p[N];

int main() {
    scanf("%d", &t);
    while (t--) {
        memset(cnt, 0, sizeof(cnt));
        scanf("%d%d", &n, &d);
        int dep = 0, sum = 0;
        for (int i = 1; i <= n;) {
            int num = min(n - i + 1, 1 << dep);
            sum += dep * num, cnt[++dep] = num, i += num;
        }
        if (sum > d || n * (n - 1) / 2 < d) {
            puts("NO");
            continue;
        }
        puts("YES");
        while (sum < d) {
            int k = 0;
            for (int i = 1; i <= dep; i++)
                if (cnt[i] * 2 > cnt[i + 1] + 2) {
                    k = i;
                    break;
                }
            cnt[k]--, cnt[k + 1]++, sum++;
            if (k == dep)dep++;
        }
        int pos = 1;
        p[1].clear();
        p[1].push_back(1), p[1].push_back(1);
        for (int i = 2; i <= dep; i++) {
            p[i].clear();
            while (cnt[i]--) {
                fa[++pos] = p[i - 1].back(), p[i - 1].pop_back();
                p[i].push_back(pos), p[i].push_back(pos);
            }
        }
        for (int i = 2; i <= n; i++)printf("%d ", fa[i]);
        puts("");
    }
    return 0;
}