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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << " , ";
  err(++it, args...);
}
const double pi = acos(-1.0);
const int mod = 998244353;
namespace fastIO {
bool IOerror = 0;
static char obuf[20000000], *OUT = obuf;
inline char nc() {
  static char buf[20000000], *p1 = buf + 20000000, *pend = buf + 20000000;
  if (p1 == pend) {
    p1 = buf;
    pend = buf + fread(buf, 1, 20000000, stdin);
    if (pend == p1) {
      IOerror = 1;
      return -1;
    }
  }
  return *p1++;
}
inline bool blank(char ch) {
  return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
}
template <typename T>
inline int read(T& x) {
  char ch;
  while (blank(ch = nc()))
    ;
  if (IOerror) return -1;
  int ok = 0;
  if (ch == '-') ok = 1, ch = nc();
  for (x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0')
    ;
  if (ok) x = -x;
  return 1;
}
template <typename T>
void print(const T& x) {
  if (x > 9) print(x / 10);
  *OUT++ = x % 10 + '0';
}
void print() { fwrite(obuf, OUT - obuf, 1, stdout); }
}  // namespace fastIO
using namespace fastIO;
long long qp(long long b, long long n) {
  long long r = 1;
  if (n < 0) n += mod - 1;
  b %= mod;
  for (; n; n >>= 1, b = b * b % mod)
    if (n & 1) r = r * b % mod;
  return r;
}
const int maxn = 2e5 + 7;
int cnt[maxn], ans[maxn];
vector<int> G[maxn];
int dfs(int u, int h) {
  int res = h;
  assert(G[u].size() <= 2);
  for (auto v : G[u]) res += dfs(v, h + 1);
  return res;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, d;
    scanf("%d%d", &n, &d);
    for (int i = 0; i <= n; ++i) cnt[i] = 0, ans[i] = 0;
    cnt[0] = 1;
    int N = n;
    int i = 1, sum = 0;
    --n;
    for (i = 1; n; ++i) {
      cnt[i] = min(n, cnt[i - 1] * 2);
      n -= cnt[i];
      sum += cnt[i] * i;
    }
    if (sum > d || N * (N - 1) / 2 < d) {
      puts("NO");
      continue;
    }
    while (1) {
      if (sum == d) break;
      for (int j = N - 1; j >= 1; --j) {
        if (cnt[j] > 1 && cnt[j + 1] + 1 <= cnt[j] * 2 - 2) {
          --cnt[j];
          ++cnt[j + 1];
          ++sum;
          break;
        }
      }
    }
    puts("YES");
    int pre = 0;
    for (int j = 1; j <= N; ++j) {
      for (int k = 1; k <= cnt[j]; ++k)
        ans[pre + cnt[j - 1] + k] = (k + 1) / 2 + pre;
      pre += cnt[j - 1];
    }
    for (int j = 2; j <= N; ++j) printf("%d%c", ans[j], " \n"[j == N]);
  }
}