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;
  mt19937 rng(
      (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
  pair<long long, long long> inv_help(long long a, long long b) {
    pair<long long, long long> res;
    res.first = 1;
    res.second = 0;
    if (b == 0)
      return res;
    else {
      pair<long long, long long> res1 = inv_help(b, a % b);
      res.first = res1.second;
      res.second = res1.first - (a / b) * res1.second;
      return res;
    }
  }
  long long inv(long long a, long long b) {
    pair<long long, long long> res = inv_help(a, b);
    long long x = res.first;
    if (x > 0) x = x % b;
    if (x < 0) x = (b - (-x) % b) % b;
    return x;
  }
  vector<int> l;
  int N;
  int binsearch(int p, int l_bnd) {
    int i = l_bnd, j = N - 1, k;
    if (l_bnd > N - 1) return -1;
    if (l[N - 1] < p)
      return -1;
    else if (l[i] > p)
      return i;
    else {
      while (i != j) {
        k = (i + j) / 2;
        if (i == j) {
          return i;
        } else if (i == j - 1) {
          if (l[i] == p)
            return i;
          else
            return i + 1;
        } else if (l[k] < p) {
          i = k + 1;
        } else if (l[k] >= p) {
          j = k;
        }
      }
      return i;
    }
  }
  vector<long long> fact(2000001, 1);
  long long nCr(int n, int r) {
    if (n < 0 || r < 0) return 0;
    if (n < r) return 0;
    long long a1 = fact[n];
    long long a2 = fact[n - r];
    long long a3 = fact[r];
    long long ans =
        (((a1 * inv(a2, 998244353)) % 998244353) * inv(a3, 998244353)) %
        998244353;
    return ans;
  }
  bool isPrime(int n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (int i = 5; i * i <= n; i = i + 6)
      if (n % i == 0 || n % (i + 2) == 0) return false;
    return true;
  }
  int sorted(vector<int>& v, int n) {
    for (int i = 1; i <= n - 1; i++) {
      if (v[i] < v[i - 1]) return 0;
    }
    return 1;
  }
  long long pow(int n) {
    if (n == 0) return 1;
    long long k = pow(n / 2);
    if (n % 2 == 1)
      return (((2 * k) % 998244353) * k) % 998244353;
    else
      return (k * k) % 998244353;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int T = 1;
    cin >> T;
    int tt = T;
    for (int i = 1; i <= 2000000; i++) fact[i] = (fact[i - 1] * i) % 998244353;
    while (T--) {
      int n;
      cin >> n;
      int ans = n;
      n++;
      ans -= n / 10;
      cout << n - ans - 1 << "\n";
    }
  }
  