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 = 1e9 + 7;
  const long long INF = 1e18;
  const long long base = 311;
  long long gcd(long long a, long long b) {
    while (b != 0) {
      long long r = a % b;
      a = b;
      b = r;
    }
    return a;
  }
  long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
  long long binmul(long long a, long long b) {
    if (b == 0) return 0;
    long long v = binmul(a, b / 2);
    long long res = v + v;
    if (res >= mod) res -= mod;
    if (b % 2) {
      res += a;
      if (res >= mod) res -= mod;
    }
    return res;
  }
  long long binpow(long long a, long long b) {
    if (b == 0) return 1;
    long long v = binpow(a, b / 2);
    if (b % 2) return v * v % mod * a % mod;
    return v * v % mod;
  }
  long long bpowmul(long long a, long long b) {
    if (b == 0) return 1;
    long long v = bpowmul(a, b / 2);
    if (b % 2) return binmul(binmul(v, v), a);
    return binmul(v, v);
  }
  const int MaxN = 2e5 + 4;
  long long n;
  void solve() {
    cin >> n;
    cout << (n + 1) / 10 << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int test;
    cin >> test;
    while (test--) solve();
    return 0;
  }
  