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;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long powmod(long long a, long long b, long long MOD) {
  if (a == 0ll) {
    return 0ll;
  }
  a %= MOD;
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = ans * a % MOD;
    }
    a = a * a % MOD;
    b >>= 1;
  }
  return ans;
}
void Pv(const vector<int> &V) {
  int Len = int(V.size());
  for (int i = 0; i < Len; ++i) {
    printf("%d", V[i]);
    if (i != Len - 1) {
      printf(" ");
    } else {
      printf("\n");
    }
  }
}
void Pvl(const vector<long long> &V) {
  int Len = int(V.size());
  for (int i = 0; i < Len; ++i) {
    printf("%lld", V[i]);
    if (i != Len - 1) {
      printf(" ");
    } else {
      printf("\n");
    }
  }
}
inline long long readll() {
  long long tmp = 0, fh = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar();
  return tmp * fh;
}
inline int readint() {
  int tmp = 0, fh = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar();
  return tmp * fh;
}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
int n, d;
int son[maxn][2];
int dep[maxn];
int fa[maxn];
int we;
std::vector<int> v[maxn];
int dfs(int x, int step, int pre) {
  v[step].push_back(x);
  fa[x] = pre;
  int res = step;
  dep[x] = step;
  if (son[x][0]) {
    res += dfs(son[x][0], step + 1, x);
  }
  if (son[x][1]) {
    res += dfs(son[x][1], step + 1, x);
  }
  if (pre == 0) we = res;
  return res;
}
void init() {
  for (int i = 0; i <= n + 2; i++) {
    son[i][0] = son[i][1] = 0;
    dep[i] = 0;
    fa[i] = 0;
  }
}
int main() {
  int t;
  t = readint();
  while (t--) {
    n = readint();
    d = readint();
    init();
    long long m2 = n - 1;
    m2 = (m2 + 1) * (m2) / 2;
    long long m1 = 0ll;
    int now = 2;
    int cnt = 1;
    for (int i = 2; i <= n; i++) {
      if (i + now - 1 <= n) {
        i += now - 1;
        m1 += 1ll * now * cnt;
        now <<= 1;
        cnt++;
      } else {
        m1 += 1ll * (n - i + 1) * cnt;
        break;
      }
    }
    if (d < m1 || d > m2) {
      printf("NO\n");
    } else {
      for (int i = 2; i <= n; i++) {
        son[i - 1][0] = i;
      }
      int id = n;
      while (dfs(1, 0, 0) != d) {
        int flag = 1;
        int j = dep[id] - 2;
        for (auto &i : v[j]) {
          if (son[i][0] == 0) {
            if (son[fa[id]][0] == id) {
              son[fa[id]][0] = 0;
            }
            if (son[fa[id]][1] == id) {
              son[fa[id]][1] = 0;
            }
            son[i][0] = id;
            flag = 0;
            break;
          } else if (son[i][1] == 0) {
            if (son[fa[id]][0] == id) {
              son[fa[id]][0] = 0;
            }
            if (son[fa[id]][1] == id) {
              son[fa[id]][1] = 0;
            }
            son[i][1] = id;
            flag = 0;
            break;
          }
        }
        if (flag) {
          id--;
        }
        for (int i = 0; i <= n; i++) {
          v[i].clear();
        }
      }
      printf("YES\n");
      for (int i = 2; i <= n; i++) {
        printf("%d%c", fa[i], i == n ? '\n' : ' ');
      }
    }
  }
  return 0;
}