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;
  template <class A, class B>
  void tostring(pair<A, B> p) {
    cerr << "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  }
  template <class A>
  void tostring(A v) {
    int f = 1;
    string r = "{";
    for (auto& x : v) {
      if (!f) {
        r += ", ";
      }
      f = 0;
      r += to_string(x);
    }
    r += "}\n";
    cerr << r;
  }
  void _out() { cerr << '\n'; }
  template <typename H, typename... T>
  void _out(H h, T... t) {
    cerr << ' ' << h;
    _out(t...);
  }
  void show() { cerr << '\n'; }
  template <typename H, typename... T>
  void show(H h, T... t) {
    cerr << h << ' ';
    show(t...);
  }
  void solve(int tase) {
    int n;
    cin >> n;
    if (n % 10 < 9)
      cout << n / 10 << '\n';
    else
      cout << n / 10 + 1 << '\n';
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T;
    cin >> T;
    for (int i = (int)(1); i <= (int)(T); i++) solve(i);
  }
  