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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  const double eps = 1e-12;
  const double PI = acos(-1.0);
  const long long INF = 1e18;
  const long long MOD = 1e9 + 7;
  const long long nax = 5e5 + 5;
  long long int gcd(long long int a, long long int b) {
    if (a == 0) return b;
    return gcd(b % a, a);
  }
  bool isPowerOfTwo(long long int x) { return x && (!(x & (x - 1))); }
  void solve() {
    long long int i, n, ans, k, m, j;
    cin >> n;
    if (n % 10 == 9) {
      ans = 1;
      n /= 10;
      n += 1;
      cout << n << "\n";
    } else {
      cout << n / 10 << "\n";
    }
    return;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ;
    long long int t = 1, i;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  