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.
#include <bits/stdc++.h>
using namespace std;
mt19937_64 rang(
    chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
  uniform_int_distribution<int> uid(0, lim - 1);
  return uid(rang);
}
int mpow(int base, int exp);
void ipgraph(int n, int m);
void dfs(int u, int par);
const int mod = 1000000007;
const int N = 3e5, M = N;
vector<int> g[N];
int a[N];
void solve() {
  long long i, j, n, k, data, m, f = 1;
  cin >> n;
  if (n == 9) {
    cout << "1\n";
    return;
  }
  data = n / 10;
  cout << data << "\n";
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
int mpow(int base, int exp) {
  base %= mod;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((long long)result * base) % mod;
    base = ((long long)base * base) % mod;
    exp >>= 1;
  }
  return result;
}
void ipgraph(int n, int m) {
  int i, u, v;
  while (m--) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
}
void dfs(int u, int par) {
  for (int v : g[u]) {
    if (v == par) continue;
    dfs(v, u);
  }
}
