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 long long MOD = 1000000007;
  const long long maxN = 1e5 + 5;
  void debug(vector<long long> vec) {
    long long n = vec.size();
    for (long long i = 0; i < n; i++) cout << vec[i] << " ";
    cout << endl;
    return;
  }
  long long gcd(long long a, long long b) {
    if (b > a) {
      return gcd(b, a);
    }
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  long long power(long long x, long long y, long long p) {
    long long res = 1;
    x = x % p;
    if (x == 0) return 0;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
  void solve() {
    long long n;
    cin >> n;
    long long ans = n / 10;
    if (n % 10 == 9) {
      cout << ans + 1 << endl;
    } else {
      cout << ans << endl;
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    long long t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  