#include <bits/stdc++.h>
using namespace std;
long long power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
int a[1001] = {0};
void hack(int pp) {
  int n;
  cin >> n;
  if (a[0] == 0) {
    int t = 0;
    int k = 1;
    while (t <= 1000) {
      if (k % 3 != 0 & (k % 10) != 3) {
        a[t] = k;
        t++;
      }
      k++;
    }
  }
  cout << a[n - 1] << "\n";
}
int main() {
  int t = 1;
  cin >> t;
  int l = 0;
  while (t--) {
    hack(l);
    l++;
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
long long power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
int a[1001] = {0};
void hack(int pp) {
  int n;
  cin >> n;
  if (a[0] == 0) {
    int t = 0;
    int k = 1;
    while (t <= 1000) {
      if (k % 3 != 0 & (k % 10) != 3) {
        a[t] = k;
        t++;
      }
      k++;
    }
  }
  cout << a[n - 1] << "\n";
}
int main() {
  int t = 1;
  cin >> t;
  int l = 0;
  while (t--) {
    hack(l);
    l++;
  }
  return 0;
}

```