#include <bits/stdc++.h>
using namespace std;
void print_ans(bool f) {
  if (f) {
    cout << "YES" << endl;
  } else {
    cout << "NO" << endl;
  }
}
long long int modular_pow(long long int base, long long int exponent,
                          long long int modulus) {
  long long int result = 1;
  while (exponent > 0) {
    if (exponent % 2 == 1) result = (result * base) % modulus;
    exponent = exponent >> 1;
    base = (base * base) % modulus;
  }
  return result;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    if (n % (10) == 9) {
      cout << n / 10 + 1 << endl;
    } else {
      cout << n / 10 << endl;
    }
  }
}

### Prompt

Generate a CPP solution to 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>
using namespace std;
void print_ans(bool f) {
  if (f) {
    cout << "YES" << endl;
  } else {
    cout << "NO" << endl;
  }
}
long long int modular_pow(long long int base, long long int exponent,
                          long long int modulus) {
  long long int result = 1;
  while (exponent > 0) {
    if (exponent % 2 == 1) result = (result * base) % modulus;
    exponent = exponent >> 1;
    base = (base * base) % modulus;
  }
  return result;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    if (n % (10) == 9) {
      cout << n / 10 + 1 << endl;
    } else {
      cout << n / 10 << endl;
    }
  }
}

```