#include <bits/stdc++.h>
using namespace std;
vector<long long> v3;
vector<long long> v1[100001];
vector<long long> v2[100001];
long long c1[100001];
long long c2[100001];
vector<pair<long long, long long>> v5;
long long sum1;
long long k1;
long long k3 = -1;
int main() {
  long long i, t;
  cin >> t;
  for (i = 0; i < t; i++) {
    long long k;
    cin >> k;
    long long l1 = 1;
    long long t1 = 0;
    long long k1 = 0;
    while (l1 <= 2000) {
      if (!(l1 % 3 == 0 || l1 % 10 == 3)) {
        t1++;
        if (t1 == k) {
          k1 = l1;
          break;
        }
      }
      l1++;
    }
    cout << k1 << endl;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
vector<long long> v3;
vector<long long> v1[100001];
vector<long long> v2[100001];
long long c1[100001];
long long c2[100001];
vector<pair<long long, long long>> v5;
long long sum1;
long long k1;
long long k3 = -1;
int main() {
  long long i, t;
  cin >> t;
  for (i = 0; i < t; i++) {
    long long k;
    cin >> k;
    long long l1 = 1;
    long long t1 = 0;
    long long k1 = 0;
    while (l1 <= 2000) {
      if (!(l1 % 3 == 0 || l1 % 10 == 3)) {
        t1++;
        if (t1 == k) {
          k1 = l1;
          break;
        }
      }
      l1++;
    }
    cout << k1 << endl;
  }
  return 0;
}

```