#include <bits/stdc++.h>
using namespace std;
class ColoredDebugCout {
 private:
  string color_code;

 public:
  ColoredDebugCout(string color_code) { this->color_code = color_code; }
  template <class T>
  void operator<<(T str) {
    cout << "\033[1;" << this->color_code << "m" << str << "\033[0m";
  }
};
auto& mcout{cout};
auto& ycout{cout};
long long s(long long x) {
  long long digit_sum{0};
  while (x > 0) {
    digit_sum += x % 10;
    x /= 10;
  }
  return digit_sum;
}
void answer1() {
  long long n;
  cin >> n;
  ycout << (n + 1) / 10;
}
void answer2() {}
void answer3() {}
void answer4() {}
void answer5() {}
void answer6() {}
void answer7() {}
void answer8() {}
void answer9() {}
int main() {
  long long t;
  cin >> t;
  for (long long test{0}; test < t; ++test) {
    answer1();
    ycout << '\n';
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
class ColoredDebugCout {
 private:
  string color_code;

 public:
  ColoredDebugCout(string color_code) { this->color_code = color_code; }
  template <class T>
  void operator<<(T str) {
    cout << "\033[1;" << this->color_code << "m" << str << "\033[0m";
  }
};
auto& mcout{cout};
auto& ycout{cout};
long long s(long long x) {
  long long digit_sum{0};
  while (x > 0) {
    digit_sum += x % 10;
    x /= 10;
  }
  return digit_sum;
}
void answer1() {
  long long n;
  cin >> n;
  ycout << (n + 1) / 10;
}
void answer2() {}
void answer3() {}
void answer4() {}
void answer5() {}
void answer6() {}
void answer7() {}
void answer8() {}
void answer9() {}
int main() {
  long long t;
  cin >> t;
  for (long long test{0}; test < t; ++test) {
    answer1();
    ycout << '\n';
  }
}

```