#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();
  }
}

### Prompt

Generate a Cpp solution to the following problem:

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.

### Response

```cpp
#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();
  }
}

```