#include <bits/stdc++.h>
using namespace std;
const int N = 1e6;
const long long INF = 1e18;
const long double EPS = 1e-9;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long random(long long a, long long b) {
  return uniform_int_distribution<long long>(a, b)(rng);
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long tt;
  cin >> tt;
  vector<long long> a;
  long long i = 1;
  while ((int)a.size() < 1000) {
    if (i % 3 != 0 && i % 10 != 3) {
      a.push_back(i);
    }
    i++;
  }
  while (tt--) {
    long long k;
    cin >> k;
    cout << a[k - 1] << '\n';
  }
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

Polycarp doesn't like integers that are divisible by 3 or end with the digit 3 in their decimal representation. Integers that meet both conditions are disliked by Polycarp, too.

Polycarp starts to write out the positive (greater than 0) integers which he likes: 1, 2, 4, 5, 7, 8, 10, 11, 14, 16, .... Output the k-th element of this sequence (the elements are numbered from 1).

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

Each test case consists of one line containing one integer k (1 ≤ k ≤ 1000).

Output

For each test case, output in a separate line one integer x — the k-th element of the sequence that was written out by Polycarp.

Example

Input


10
1
2
3
4
5
6
7
8
9
1000


Output


1
2
4
5
7
8
10
11
14
1666

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6;
const long long INF = 1e18;
const long double EPS = 1e-9;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long random(long long a, long long b) {
  return uniform_int_distribution<long long>(a, b)(rng);
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long tt;
  cin >> tt;
  vector<long long> a;
  long long i = 1;
  while ((int)a.size() < 1000) {
    if (i % 3 != 0 && i % 10 != 3) {
      a.push_back(i);
    }
    i++;
  }
  while (tt--) {
    long long k;
    cin >> k;
    cout << a[k - 1] << '\n';
  }
  return 0;
}

```