## Algorithm

B. Make it Divisible by 25
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

It is given a positive integer n. In 11 move, one can select any single digit and remove it (i.e. one selects some position in the number and removes the digit located at this position). The operation cannot be performed if only one digit remains. If the resulting number contains leading zeroes, they are automatically removed.

E.g. if one removes from the number 3292532925 the 33-rd digit, the resulting number will be 32253225. If one removes from the number 2009905020099050 the first digit, the resulting number will be 9905099050 (the 22 zeroes going next to the first digit are automatically removed).

What is the minimum number of steps to get a number such that it is divisible by 2525 and positive? It is guaranteed that, for each n occurring in the input, the answer exists. It is guaranteed that the number n has no leading zeros.

Input

The first line contains one integer t (1t1041≤�≤104) — the number of test cases. Then t test cases follow.

Each test case consists of one line containing one integer n (25n101825≤�≤1018). It is guaranteed that, for each n occurring in the input, the answer exists. It is guaranteed that the number n has no leading zeros.

Output

For each test case output on a separate line an integer k (k0�≥0) — the minimum number of steps to get a number such that it is divisible by 2525 and positive.

Example
input
Copy
5
100
71345
3259
50555
2050047
output
Copy
0
3
1
3
2
Note

In the first test case, it is already given a number divisible by 2525.

In the second test case, we can remove the digits 1133, and 44 to get the number 7575.

In the third test case, it's enough to remove the last digit to get the number 325325.

In the fourth test case, we can remove the three last digits to get the number 5050.

In the fifth test case, it's enough to remove the digits 44 and 77.

## Code Examples

### #1 Code Example with C++ Programming

Code - C++ Programming

#include<bits/stdc++.h>
using namespace std;

#define ll long long
#define endl '\n'
#define debug(n) cout<<(n)<<endl;
const ll INF = 2e18 + 99;

int totcount(string s, string t) {
int count = 0;
int l = s.length();
while(s.length() > 0 && t.length() > 0) {
if(s.back() == t.back()) {
t.pop_back();
}
else {
count++;
}
s.pop_back();
}
if(t.length() == 0){
return count;
}
return l;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin>>t;
while(t--) {
string s;
cin>>s;
if(s.length() < 2) {
cout << s.length()<<endl;
} else {
cout<<min({totcount(s,"00"),totcount(s,"25"),totcount(s,"50"),totcount(s,"75")})<<endl;
}
}
}
Copy The Code &

Input

cmd
5
100
71345
3259
50555
2050047

Output

cmd
0
3
1
3
2