#include <bits/stdc++.h>
using namespace std;
int isprime(long long num) {
  if (num == 2) return 1;
  if (num <= 1 || num % 2 == 0) return 0;
  for (long long i = 3; i * i <= num; i += 2) {
    if (num % i == 0) return 0;
  }
  return 1;
}
long long sum1(long long n) {
  long long cnt = 0;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      cnt++;
      if (n / i != i) cnt++;
    }
  }
  return cnt;
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    int num;
    cin >> num;
    num++;
    cout << num / 10 << "\n";
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
int isprime(long long num) {
  if (num == 2) return 1;
  if (num <= 1 || num % 2 == 0) return 0;
  for (long long i = 3; i * i <= num; i += 2) {
    if (num % i == 0) return 0;
  }
  return 1;
}
long long sum1(long long n) {
  long long cnt = 0;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      cnt++;
      if (n / i != i) cnt++;
    }
  }
  return cnt;
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    int num;
    cin >> num;
    num++;
    cout << num / 10 << "\n";
  }
}

```