#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  int ans[t];
  for (int l = 0; l < t; l++) {
    int k, n, i;
    cin >> k;
    i = k % 18;
    if (i == 1)
      n = 1;
    else if (i == 2)
      n = 2;
    else if (i == 3)
      n = 4;
    else if (i == 4)
      n = 5;
    else if (i == 5)
      n = 7;
    else if (i == 6)
      n = 8;
    else if (i == 7)
      n = 10;
    else if (i == 8)
      n = 11;
    else if (i == 9)
      n = 14;
    else if (i == 10)
      n = 16;
    else if (i == 11)
      n = 17;
    else if (i == 12)
      n = 19;
    else if (i == 13)
      n = 20;
    else if (i == 14)
      n = 22;
    else if (i == 15)
      n = 25;
    else if (i == 16)
      n = 26;
    else if (i == 17)
      n = 28;
    else if (i == 0) {
      n = 29;
      k--;
    }
    ans[l] = (30 * (k / 18)) + n;
  }
  for (int l = 0; l < t; l++) {
    cout << ans[l] << endl;
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
int main() {
  int t;
  cin >> t;
  int ans[t];
  for (int l = 0; l < t; l++) {
    int k, n, i;
    cin >> k;
    i = k % 18;
    if (i == 1)
      n = 1;
    else if (i == 2)
      n = 2;
    else if (i == 3)
      n = 4;
    else if (i == 4)
      n = 5;
    else if (i == 5)
      n = 7;
    else if (i == 6)
      n = 8;
    else if (i == 7)
      n = 10;
    else if (i == 8)
      n = 11;
    else if (i == 9)
      n = 14;
    else if (i == 10)
      n = 16;
    else if (i == 11)
      n = 17;
    else if (i == 12)
      n = 19;
    else if (i == 13)
      n = 20;
    else if (i == 14)
      n = 22;
    else if (i == 15)
      n = 25;
    else if (i == 16)
      n = 26;
    else if (i == 17)
      n = 28;
    else if (i == 0) {
      n = 29;
      k--;
    }
    ans[l] = (30 * (k / 18)) + n;
  }
  for (int l = 0; l < t; l++) {
    cout << ans[l] << endl;
  }
  return 0;
}

```