#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx")
using namespace std;
using namespace chrono;
const int MOD = (1e9 + 7);
long long power(long long a, long long b) {
  long long res = 1;
  a = a % MOD;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % MOD;
      b--;
    }
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t, n, i, j, k, f;
  cin >> t;
  for (int tc = 1; tc <= t; tc++) {
    cin >> n;
    cout << n / 10 + bool(n % 10 == 9) << "\n";
  }
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

Let's define S(x) to be the sum of digits of number x written in decimal system. For example, S(5) = 5, S(10) = 1, S(322) = 7.

We will call an integer x interesting if S(x + 1) < S(x). In each test you will be given one integer n. Your task is to calculate the number of integers x such that 1 ≤ x ≤ n and x is interesting.

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — number of test cases.

Then t lines follow, the i-th line contains one integer n (1 ≤ n ≤ 10^9) for the i-th test case.

Output

Print t integers, the i-th should be the answer for the i-th test case.

Example

Input


5
1
9
10
34
880055535


Output


0
1
1
3
88005553

Note

The first interesting number is equal to 9.

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx")
using namespace std;
using namespace chrono;
const int MOD = (1e9 + 7);
long long power(long long a, long long b) {
  long long res = 1;
  a = a % MOD;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % MOD;
      b--;
    }
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t, n, i, j, k, f;
  cin >> t;
  for (int tc = 1; tc <= t; tc++) {
    cin >> n;
    cout << n / 10 + bool(n % 10 == 9) << "\n";
  }
  return 0;
}

```