#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
using ll = long long;
using datas = pair<ll, ll>;
using ddatas = pair<long double, long double>;
using tdata = pair<ll, datas>;
using vec = vector<ll>;
using mat = vector<vec>;
using pvec = vector<datas>;
using pmat = vector<pvec>;
const ll mod = 1000000007;
const ll inf = 1LL << 60;
const long double PI = acos(-1);
const long double eps = 1e-9;
template <class T>
inline bool chmax(T& a, T b) {
  bool x = a < b;
  if (x) a = b;
  return x;
}
template <class T>
inline bool chmin(T& a, T b) {
  bool x = a > b;
  if (x) a = b;
  return x;
}
void startupcpp() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);
}
long double distance(ddatas x, ddatas y) {
  long double a = x.first - y.first, b = x.second - y.second;
  return sqrt(a * a + b * b);
}
ll modinv(ll a) {
  ll b = mod, u = 1, v = 0, t;
  while (b) {
    t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  return (u + mod) % mod;
}
ll moddevide(ll a, ll b) { return (a * modinv(b)) % mod; }
vec modncrlistp, modncrlistm;
ll modncr(ll n, ll r) {
  ll i, size = modncrlistp.size();
  if (size <= n) {
    modncrlistp.resize(n + 1);
    modncrlistm.resize(n + 1);
    if (!size) {
      modncrlistp[0] = modncrlistm[0] = 1;
      size++;
    }
    for (i = size; i < (ll)n + 1; i++) {
      modncrlistp[i] = modncrlistp[i - 1] * i % mod;
      modncrlistm[i] = modinv(modncrlistp[i]);
    }
  }
  return modncrlistp[n] * modncrlistm[r] % mod * modncrlistm[n - r] % mod;
}
ll modpow(ll a, ll n) {
  ll res = 1;
  while (n) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}
ll gcd(ll a, ll b) {
  if (!b) return a;
  return (a % b == 0) ? b : gcd(b, a % b);
}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll countdigits(ll n) {
  ll ans = 0;
  while (n) {
    n /= 10;
    ans++;
  }
  return ans;
}
ll sumdigits(ll n) {
  ll ans = 0;
  while (n) {
    ans += n % 10;
    n /= 10;
  }
  return ans;
}
int main() {
  startupcpp();
  ll test;
  cin >> test;
  while (test--) {
    ll i, j = 1, N, D, mindis = 0;
    cin >> N >> D;
    vec dist = {1, 0}, ansdis(N, 0), lim(N, 2), ans(N - 1);
    for (i = 1; i < (ll)N; i++) {
      dist[j]++;
      mindis += j;
      if (dist[j] == dist[j - 1] * 2) {
        j++;
        if (i != N - 1) dist.push_back(0);
      }
    }
    if (mindis > D || D > N * (N - 1) / 2) {
      cout << "NO"
           << "\n";
      continue;
    }
    cout << "YES"
         << "\n";
    i = j = (int)dist.size() - 1;
    while (mindis < D) {
      ++mindis;
      if (dist.back() > 1) {
        dist[(int)dist.size() - 1]--;
        dist.push_back(1);
        continue;
      }
      if (dist[j] == 1) {
        while (dist[i] == 1) i--;
        j = i;
      }
      dist[j]--;
      dist[++j]++;
    }
    j = 0;
    for (i = 0; i < (ll)N; i++) {
      ansdis[i] = j;
      if (--dist[j] == 0) j++;
    }
    j = 0;
    for (i = 1; i < (ll)N; i++) {
      while (ansdis[i] != ansdis[j] + 1 || !lim[j]) j++;
      ans[i - 1] = j + 1;
      lim[j]--;
    }
    do {
      bool f = 0;
      for (auto outi : ans) {
        cout << (f ? " " : "") << outi;
        f = 1;
      }
      cout << "\n";
    } while (0);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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>

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
using ll = long long;
using datas = pair<ll, ll>;
using ddatas = pair<long double, long double>;
using tdata = pair<ll, datas>;
using vec = vector<ll>;
using mat = vector<vec>;
using pvec = vector<datas>;
using pmat = vector<pvec>;
const ll mod = 1000000007;
const ll inf = 1LL << 60;
const long double PI = acos(-1);
const long double eps = 1e-9;
template <class T>
inline bool chmax(T& a, T b) {
  bool x = a < b;
  if (x) a = b;
  return x;
}
template <class T>
inline bool chmin(T& a, T b) {
  bool x = a > b;
  if (x) a = b;
  return x;
}
void startupcpp() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);
}
long double distance(ddatas x, ddatas y) {
  long double a = x.first - y.first, b = x.second - y.second;
  return sqrt(a * a + b * b);
}
ll modinv(ll a) {
  ll b = mod, u = 1, v = 0, t;
  while (b) {
    t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  return (u + mod) % mod;
}
ll moddevide(ll a, ll b) { return (a * modinv(b)) % mod; }
vec modncrlistp, modncrlistm;
ll modncr(ll n, ll r) {
  ll i, size = modncrlistp.size();
  if (size <= n) {
    modncrlistp.resize(n + 1);
    modncrlistm.resize(n + 1);
    if (!size) {
      modncrlistp[0] = modncrlistm[0] = 1;
      size++;
    }
    for (i = size; i < (ll)n + 1; i++) {
      modncrlistp[i] = modncrlistp[i - 1] * i % mod;
      modncrlistm[i] = modinv(modncrlistp[i]);
    }
  }
  return modncrlistp[n] * modncrlistm[r] % mod * modncrlistm[n - r] % mod;
}
ll modpow(ll a, ll n) {
  ll res = 1;
  while (n) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}
ll gcd(ll a, ll b) {
  if (!b) return a;
  return (a % b == 0) ? b : gcd(b, a % b);
}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll countdigits(ll n) {
  ll ans = 0;
  while (n) {
    n /= 10;
    ans++;
  }
  return ans;
}
ll sumdigits(ll n) {
  ll ans = 0;
  while (n) {
    ans += n % 10;
    n /= 10;
  }
  return ans;
}
int main() {
  startupcpp();
  ll test;
  cin >> test;
  while (test--) {
    ll i, j = 1, N, D, mindis = 0;
    cin >> N >> D;
    vec dist = {1, 0}, ansdis(N, 0), lim(N, 2), ans(N - 1);
    for (i = 1; i < (ll)N; i++) {
      dist[j]++;
      mindis += j;
      if (dist[j] == dist[j - 1] * 2) {
        j++;
        if (i != N - 1) dist.push_back(0);
      }
    }
    if (mindis > D || D > N * (N - 1) / 2) {
      cout << "NO"
           << "\n";
      continue;
    }
    cout << "YES"
         << "\n";
    i = j = (int)dist.size() - 1;
    while (mindis < D) {
      ++mindis;
      if (dist.back() > 1) {
        dist[(int)dist.size() - 1]--;
        dist.push_back(1);
        continue;
      }
      if (dist[j] == 1) {
        while (dist[i] == 1) i--;
        j = i;
      }
      dist[j]--;
      dist[++j]++;
    }
    j = 0;
    for (i = 0; i < (ll)N; i++) {
      ansdis[i] = j;
      if (--dist[j] == 0) j++;
    }
    j = 0;
    for (i = 1; i < (ll)N; i++) {
      while (ansdis[i] != ansdis[j] + 1 || !lim[j]) j++;
      ans[i - 1] = j + 1;
      lim[j]--;
    }
    do {
      bool f = 0;
      for (auto outi : ans) {
        cout << (f ? " " : "") << outi;
        f = 1;
      }
      cout << "\n";
    } while (0);
  }
  return 0;
}

```