Problem: Vasya came up with a password to register for EatForces — a string s. 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 s is a string x = s_l s_{l + 1} ... s_{l + len - 1} (1 ≤ l ≤ |s|, 0 ≤ len ≤ |s| - l + 1). len is the length of the substring. Note that the empty string is also considered a substring of s, it has the length 0.

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 exactly once, and the chosen string should have the minimal possible length.

Note that the length of s 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 T (1 ≤ T ≤ 100) — the number of testcases.

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

Only T = 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 0. 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 4, because the changed positions are 2 and 5, thus (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

2
abcDCE
htQw27


Output

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).

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class VasyaAndPassword {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader inp = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Solver solver = new Solver();
        solver.solve(inp, out);
        out.close();
    }

    static class Solver {
        private int best = Integer.MAX_VALUE;
        private String res = "";

        private boolean low(char a) {
            return 97 <= a && a <= 122;
        }
        private boolean high(char a) {
            return 65 <= a && a <= 90;
        }
        private boolean digit(char a) {
            return !(low(a) || high(a));
        }
        private void res2(String b, int x) {
            char[] a = b.toCharArray();
            int[] cnt = new int[3];
            for (char c: a) {
                if (low(c)) cnt[0]++;
                else if (high(c)) cnt[1]++;
                else cnt[2]++;
            }
            boolean low = cnt[0] > 0;
            boolean high = cnt[1] > 0;
            boolean digit = cnt[2] > 0;
            for (int i = x; i < b.length(); i++) {
                if ((low(a[i]) && cnt[0] > 1) || (high(a[i]) && cnt[1] > 1) || (digit(a[i]) && cnt[2] > 1)) {
                    if (!low || !high || !digit) {
                        if (cnt[0] > 1) cnt[0]--;
                        else if (cnt[1] > 1) cnt[1]--;
                        else cnt[2]--;
                    }
                    if (!low) {
                        a[i] = 'a';
                        low = true;
                    }
                    else if (!high) {
                        a[i] = 'A';
                        high = true;
                    }
                    else if (!digit) {
                        a[i] = '0';
                        digit = true;
                    }
                }
                if (low && high && digit) {
                    if (i - x < best) {
                        best = i - x;
                        StringBuilder sb = new StringBuilder();
                        for (char d: a) sb.append(d);
                        res = sb.toString();
                    }
                }
            }
        }

        private void solve(InputReader inp, PrintWriter out) {
            int t = inp.nextInt();
            for (int i = 0; i < t; i++) {
                String a = inp.next();
                best = Integer.MAX_VALUE;
                for (int j = 0; j < a.length(); j++) res2(new String(a), j);
                out.println(res);
            }
        }
    }

    static class InputReader {
        BufferedReader reader;
        StringTokenizer tokenizer;

        InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }
    }
}