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>
const int inf = 0x3f3f3f3f, Inf = 0x7fffffff;
const long long INF = 0x7fffffffffffffff;
const double eps = 1e-10;
unsigned int seed = 19260817;
const unsigned int _RAND_MAX_ = 4294967295u;
__inline__ __attribute__((always_inline)) unsigned int Rand() {
  return seed = seed * 998244353u + 1000000007u;
}
template <typename _Tp>
_Tp gcd(const _Tp &a, const _Tp &b) {
  return (!b) ? a : gcd(b, a % b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp abs(const _Tp &a) {
  return a >= 0 ? a : -a;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp max(const _Tp &a, const _Tp &b) {
  return a < b ? b : a;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp min(const _Tp &a, const _Tp &b) {
  return a < b ? a : b;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void chmax(_Tp &a, const _Tp &b) {
  (a < b) && (a = b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void chmin(_Tp &a, const _Tp &b) {
  (b < a) && (a = b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) bool _cmp(const _Tp &a,
                                                    const _Tp &b) {
  return abs(a - b) <= eps;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void read(_Tp &x) {
  register char ch(getchar());
  bool f(false);
  while (ch < 48 || ch > 57) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (ch >= 48 && ch <= 57)
    x = (((x << 2) + x) << 1) + (ch & 15), ch = getchar();
  if (f) x = -x;
}
template <typename _Tp, typename... Args>
__inline__ __attribute__((always_inline)) void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
__inline__ __attribute__((always_inline)) int read_str(char *s) {
  register char ch(getchar());
  while (ch == ' ' || ch == '\r' || ch == '\n') ch = getchar();
  register char *tar = s;
  *tar = ch, ch = getchar();
  while (ch != ' ' && ch != '\r' && ch != '\n' && ch != EOF)
    *(++tar) = ch, ch = getchar();
  return tar - s + 1;
}
const int N = 5005;
long long cnt[N];
void init() {
  for (int i = 1; i < N; ++i) {
    cnt[i] = cnt[(i - 1) >> 1] + cnt[(i - 1) - ((i - 1) >> 1)] + i - 1;
  }
}
int pos;
int fa[N];
__inline__ __attribute__((always_inline)) bool check(int n, int d) {
  return cnt[n] <= d && n * (n - 1) / 2 >= d;
}
__inline__ __attribute__((always_inline)) bool check(int n1, int n2, int d) {
  return cnt[n1] + cnt[n2] <= d && n1 * (n1 - 1) / 2 + n2 * (n2 - 1) / 2 >= d;
}
int solve(int n, int d) {
  if (!n) return 0;
  d -= n - 1;
  int cur = ++pos;
  int sizl = 0, sizr = 0;
  for (int i = n - 1; i >= 0; --i) {
    if (check(i, n - i - 1, d)) {
      sizl = i, sizr = n - i - 1;
      break;
    }
  }
  for (int i = 0; i <= d; ++i) {
    if (check(sizl, i) && check(sizr, d - i)) {
      fa[solve(sizl, i)] = cur;
      fa[solve(sizr, d - i)] = cur;
      break;
    }
  }
  return cur;
}
void MAIN() {
  int n, d;
  read(n, d);
  if (!check(n, d)) {
    printf("NO\n");
    return;
  }
  pos = 0;
  solve(n, d);
  printf("YES\n");
  for (int i = 2; i <= n; ++i) printf("%d ", fa[i]);
  printf("\n");
}
int main() {
  init();
  int _;
  read(_);
  while (_--) MAIN();
  return 0;
}