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 <typename Arg1>
  void __f(const char* name, Arg1&& arg1) {
    std::cerr << name << " : " << arg1 << endl;
  }
  template <typename Arg1, typename... Args>
  void __f(const char* names, Arg1&& arg1, Args&&... args) {
    const char* comma = strchr(names + 1, ',');
    std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
    __f(comma + 1, args...);
  }
  class A {
    int x;
  
   public:
    A(int i) { x = i; }
    void print() { cout << x; }
  };
  class B : public A {
   public:
    B() : A(10) {}
  };
  class C : public A {
   public:
    C() : A(10) {}
  };
  class D : public B, public C {};
  void solve() {
    long long int a = 1, b = 0, c = 0, t = 0, ans2 = 0, k = -1, ans = 0, c1 = 0,
                  c2 = 0, c3 = 0, c4 = 0, n = 0, m = 0, x = 1e18;
    cin >> a;
    if (a < 9)
      cout << 0 << endl;
    else
      cout << ((a - 9) / 10 + 1) << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    clock_t z = clock();
    long long int qc = 1, additionalv;
    cin >> qc;
    for (long long int i = 1; i <= qc; i++) {
      solve();
    }
    fprintf(stderr, "Real Time: % .4Lf\n",
            (long double)(clock() - z) / CLOCKS_PER_SEC),
        fflush(stderr);
  }
  