下面是题目复述：

You are given three integers $aa$, $bb$ and $cc$.

Find two positive integers $xx$ and $yy$ ($x>0x\; >\; 0$, $y>0y\; >\; 0$) such that:

the decimal representation of $xx$ without leading zeroes consists of $aa$ digits; the decimal representation of $yy$ without leading zeroes consists of $bb$ digits; the decimal representation of $gcd(x,y)gcd(x,\; y)$ without leading zeroes consists of $cc$ digits. $gcd(x,y)gcd(x,\; y)$ denotes the greatest common divisor (GCD) of integers $xx$ and $yy$.

Output

$xx$ and $yy$. If there are multiple answers, output any of them.

Input

The first line contains a single integer $tt$ ($1\le t\le 2851\; \backslash le\; t\; \backslash le\; 285$) — the number of testcases.

Each of the next $tt$ lines contains three integers $aa$, $bb$ and $cc$ ($1\le a,b\le 91\; \backslash le\; a,\; b\; \backslash le\; 9$, $1\le c\le min(a,b)1\; \backslash le\; c\; \backslash le\; min(a,\; b)$) — the required lengths of the numbers.

It can be shown that the answer exists for all testcases under the given constraints.

Additional constraint on the input: all testcases are different.

Output

For each testcase print two positive integers — $xx$ and $yy$ ($x>0x\; >\; 0$, $y>0y\; >\; 0$) such that

the decimal representation of $xx$ without leading zeroes consists of $aa$ digits;

the decimal representation of $yy$ without leading zeroes consists of $bb$ digits;

the decimal representation of $gcd(x,y)gcd(x,\; y)$ without leading zeroes consists of $cc$ digits.

Example

input

4

2 3 1

2 2 2

6 6 2

1 1 1

output

11 492

13 26

140133 160776

1 1

Note

In the example:

$gcd(11,492)=1gcd(11,\; 492)\; =\; 1$

$gcd(13,26)=13gcd(13,\; 26)\; =\; 13$

$gcd(140133,160776)=21gcd(140133,\; 160776)\; =\; 21$

$gcd(1,1)=1gcd(1,\; 1)\; =\; 1$

下面是解题思路：

题目大意：

输入三个数字a,b,c.

• a表示数字x的位数
• b表示数字y的位数
• c表示数字gcd（x,y）的位数。

（gcd（x，y）表示x和y的最大公约数）

求出满足条件的x和y，输出一组即可。

思路：

暴力枚举，把a位数下面的所有数字和b位数下面的的所有数字枚举求最大公约数，然后得到答案。

需要注意的是，pow很可能数据丢失精确度，可以提前打表确定几位数的开始数字。

下面是AC代码：

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;
bool flag=false;
int gcd(int a,int b)
{
    return b?gcd(b,a%b):a;
}
int bei[11];
void in()
{
    bei[0]=1;
    for(int i=1; i<10; i++)
        bei[i]=bei[i-1]*10;
}
int main()
{
    in();
    int t;
    scanf("%d",&t);
    while (t--)
    {
        int a,b,c;
        scanf("%d %d %d",&a,&b,&c);
        for(int i=bei[a-1]; i<=bei[a]; i++)
        {
            for(int j=bei[b-1]; j<=bei[b]; j++)
            {
                int ans=gcd(i,j),num=0;
                for(;ans>0;ans/=10) num++;
                if(num==c)
                {
                    printf("%d %d\n",i,j);
                    flag=true;
                    break;
                }
            }
            if(flag) break;
        }

    }
    return 0;
}