A. Vasya And Password(字符串)

滴答滴答---题目链接

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya came up with a password to register for EatForces — a string ss. The password in EatForces should be a string, consisting of lowercase and uppercase Latin letters and digits.

But since EatForces takes care of the security of its users, user passwords must contain at least one digit, at least one uppercase Latin letter and at least one lowercase Latin letter. For example, the passwords "abaCABA12", "Z7q" and "3R24m" are valid, and the passwords "qwerty", "qwerty12345" and "Password" are not.

A substring of string ss is a string x=slsl+1…sl+len−1(1≤l≤|s|,0≤len≤|s|−l+1)x=slsl+1…sl+len−1(1≤l≤|s|,0≤len≤|s|−l+1). lenlen is the length of the substring. Note that the empty string is also considered a substring of ss, it has the length 00.

Vasya's password, however, may come too weak for the security settings of EatForces. He likes his password, so he wants to replace some its substring with another string of the same length in order to satisfy the above conditions. This operation should be performed exactlyonce, and the chosen string should have the minimal possible length.

Note that the length of ss should not change after the replacement of the substring, and the string itself should contain only lowercase and uppercase Latin letters and digits.

Input

The first line contains a single integer TT (1≤T≤1001≤T≤100) — the number of testcases.

Each of the next TT lines contains the initial password s (3≤|s|≤100)s (3≤|s|≤100), consisting of lowercase and uppercase Latin letters and digits.

Only T=1T=1 is allowed for hacks.

Output

For each testcase print a renewed password, which corresponds to given conditions.

The length of the replaced substring is calculated as following: write down all the changed positions. If there are none, then the length is 00. Otherwise the length is the difference between the first and the last changed position plus one. For example, the length of the changed substring between the passwords "abcdef" →→ "a7cdEf" is 44, because the changed positions are 22 and 55, thus (5−2)+1=4(5−2)+1=4.

It is guaranteed that such a password always exists.

If there are several suitable passwords — output any of them.

Example

input

Copy

2
abcDCE
htQw27

output

Copy

abcD4E
htQw27

Note

In the first example Vasya's password lacks a digit, he replaces substring "C" with "4" and gets password "abcD4E". That means, he changed the substring of length 1.

In the second example Vasya's password is ok from the beginning, and nothing has to be changed. That is the same as replacing the empty substring with another empty substring (length 0).

题目大意：给出n个字符串，字符串要满足有大写字母，小写字母，数字才能构成密码，

要求以最小改动把字符串变为密码字符串

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6;
vector <int> v[3];
int main()
{
    int T;
    cin >> T;
char s[1010];
    while(T--)
    {
        scanf("%s",s);
        for(int i=0; i<3; i++)
            v[i].clear();
            int len=strlen(s);
        for(int i=0; i<len; i++)
        {
            if(s[i]>='A'&&s[i]<='Z')
            {
                v[0].push_back(i);
            }
            else if(s[i]>='a'&&s[i]<='z')
            {
                v[1].push_back(i);
            }
            else if(s[i]>='0'&&s[i]<='9')
            {
                v[2].push_back(i);
            }
        }
        bool flag=0;
        for(int i=0; i<3; i++)
        {
            if(!v[i].size())
            {
                flag=1;
                break;
            }
        }
        if(!flag)cout <<s << endl;
        else
        {
            if(!v[0].size())
            {
                for(int i=0; i<3; i++)
                {
                    if(i==0)continue;
                    if(v[i].size()>1)
                    {
                        s[v[i][v[i].size()-1]]='A';
                        v[i].pop_back();
                        break;
                    }
                }
            }
            if(!v[1].size())
            {
                for(int i=0; i<3; i++)
                {
                    if(i==1)continue;
                    if(v[i].size()>1)
                    {
                        s[v[i][v[i].size()-1]]='a';
                        v[i].pop_back();
                        break;
                    }
                }
            }
            if(!v[2].size())
            {
                for(int i=0; i<3; i++)
                {
                    if(i==2)continue;
                    if(v[i].size()>1)
                    {
                        s[v[i][v[i].size()-1]]='1';
                        v[i].pop_back();
                        break;
                    }
                }
            }
            cout << s << endl;
        }
    }
    return 0;
}