description: 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.
solution: #include <bits/stdc++.h>
  using namespace std;
  const double pi = acos(-1.0);
  int maxar(int arr[], int n) {
    int res = arr[0];
    for (int i = 0; i < n; i++) {
      res = max(res, arr[i]);
    }
    return res;
  }
  long long maxar(long long arr[], long long n) {
    long long res = arr[0];
    for (long long i = 0; i < n; i++) {
      res = max(res, arr[i]);
    }
    return res;
  }
  long long minar(long long arr[], long long n) {
    long long res = arr[0];
    for (long long i = 0; i < n; i++) {
      res = min(res, arr[i]);
    }
    return res;
  }
  int minar(int arr[], int n) {
    int res = arr[0];
    for (int i = 0; i < n; i++) {
      res = min(res, arr[i]);
    }
    return res;
  }
  void inpar(int arr[], int n) {
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
    }
  }
  void inpar(long long arr[], long long n) {
    for (long long i = 0; i < n; i++) {
      cin >> arr[i];
    }
  }
  template <typename T>
  inline T readInt() {
    T n = 0, s = 1;
    char p = getchar();
    if (p == '-') s = -1;
    while ((p < '0' || p > '9') && p != EOF && p != '-') p = getchar();
    if (p == '-') s = -1, p = getchar();
    while (p >= '0' && p <= '9') {
      n = (n << 3) + (n << 1) + (p - '0');
      p = getchar();
    }
    return n * s;
  }
  bool isPalin(string s, int n) {
    int i = 0, j = n - 1;
    while (i < j) {
      if (s[i] != s[j]) {
        return false;
      }
      i++;
      j--;
    }
    return true;
  }
  void solve() {
    int n = readInt<int>();
    int x = (n % 10 == 9) ? 1 : 0;
    cout << n / 10 + x << endl;
  }
  int32_t main() {
    int tc;
    cin >> tc;
    while (tc--) {
      solve();
    }
    return 0;
  }
  