#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int MAX = 1000000000;
const int inf = 1e9;
const long double PI = 3.141592653589793238460;
int max(int a, int b) {
  if (a > b) return a;
  return b;
}
void solve() {
  int n;
  cin >> n;
  if (n < 9) {
    cout << 0 << '\n';
    return;
  }
  if (n == 9)
    cout << 1 << '\n';
  else {
    if (n % 10 == 9)
      cout << 1 + n / 10 << '\n';
    else
      cout << n / 10 << '\n';
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
}

### Prompt

Please create a solution in cpp 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>
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int MAX = 1000000000;
const int inf = 1e9;
const long double PI = 3.141592653589793238460;
int max(int a, int b) {
  if (a > b) return a;
  return b;
}
void solve() {
  int n;
  cin >> n;
  if (n < 9) {
    cout << 0 << '\n';
    return;
  }
  if (n == 9)
    cout << 1 << '\n';
  else {
    if (n % 10 == 9)
      cout << 1 + n / 10 << '\n';
    else
      cout << n / 10 << '\n';
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
}

```