Binary Period CodeForces

Let's say string ss has period kk if si=si+ksi=si+k for all ii from 11 to |s|−k|s|−k (|s||s| means length of string ss) and kk is the minimum positive integer with this property.

Some examples of a period: for ss="0101" the period is k=2k=2, for ss="0000" the period is k=1k=1, for ss="010" the period is k=2k=2, for ss="0011" the period is k=4k=4.

You are given string tt consisting only of 0's and 1's and you need to find such string ss that:

1. String ss consists only of 0's and 1's;
2. The length of ss doesn't exceed 2⋅|t|2⋅|t|;
3. String tt is a subsequence of string ss;
4. String ss has smallest possible period among all strings that meet conditions 1—3.

Let us recall that tt is a subsequence of ss if tt can be derived from ss by deleting zero or more elements (any) without changing the order of the remaining elements. For example, tt="011" is a subsequence of ss="10101".

Input

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

Next TT lines contain test cases — one per line. Each line contains string tt (1≤|t|≤1001≤|t|≤100) consisting only of 0's and 1's.

Output

Print one string for each test case — string ss you needed to find. If there are multiple solutions print any one of them.

Example

Input

4
00
01
111
110

Output

00
01
11111
1010

Note

In the first and second test cases, s=ts=t since it's already one of the optimal solutions. Answers have periods equal to 11 and 22, respectively.

In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string ss. String ss has period equal to 11.

水思维题，，有‘1’有‘0’就1010...走起，else 输出一列s[1]

#include <bits/stdc++.h>
using namespace std;
char s[200100];
int main() {
    int t;
    scanf("%d",&t);
    while(t--) {
        scanf("%s",s+1);
        int n=strlen(s+1);
        bool a=0,b=0;
        for (int i=1; i<=n; i++) {
            if(s[i]=='0') a=1;
            if(s[i]=='1') b=1;
        }
        if(!a||!b) {
            for (int i=1; i<=n; i++)
                            printf("%c",s[1]);
        } 
                 else {
            for (int i=1; i<=n; i++)
                            printf("10");
        }
        puts("");
    }
    return 0;
}