Problem: You are given three integers a, b and c.

Find two positive integers x and y (x > 0, y > 0) such that: 

  * the decimal representation of x without leading zeroes consists of a digits; 
  * the decimal representation of y without leading zeroes consists of b digits; 
  * the decimal representation of gcd(x, y) without leading zeroes consists of c digits. 



gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.

Output x and y. If there are multiple answers, output any of them.

Input

The first line contains a single integer t (1 ≤ t ≤ 285) — the number of testcases.

Each of the next t lines contains three integers a, b and c (1 ≤ a, b ≤ 9, 1 ≤ c ≤ 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 — x and y (x > 0, y > 0) such that 

  * the decimal representation of x without leading zeroes consists of a digits; 
  * the decimal representation of y without leading zeroes consists of b digits; 
  * the decimal representation of gcd(x, y) without leading zeroes consists of c 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: 

  1. gcd(11, 492) = 1 
  2. gcd(13, 26) = 13 
  3. gcd(140133, 160776) = 21 
  4. gcd(1, 1) = 1 

Solution (in CPP):

Solution:
//				coder name:mark_41
//						collage:mmmut
//								wannabe coder
#include <bits/stdc++.h>
 
using namespace std;
 
#define ll long long int
#define REP(i,a,b) for(ll i=a;i<b;i++)
#define RREP(i,b,a) for(ll i=b;i>=a;i--)
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define v(type) vector<type>
const ll mode= 1e17;
const ll nmode= (-1)*1e17;
#define MAX  5000001
ll gcd(ll a,ll b)
{
    if(b==0)
    {
        return a;
    }
    gcd(b,a%b);
    
}
int main(){
 ll t;
 cin>>t;
	while(t--)
	{ll a,b,c;
	cin>>a>>b>>c;
	ll s1=pow(10,c-1);
	ll mn,mx;
	if(a>b)
	{
	    mn=b;
	    mx=a;
	    	ll x1=1;
	REP(i,0,mx-c)
	{
	   x1=x1*10+1;
	}
	    
	    ll x2=pow(10,mn-1);
	    cout<<x1*s1<<" "<<x2;
	    cout<<"\n";
	}
	else{
	    mn=a;
	    mx=b;
	    	ll x1=1;
	REP(i,0,mx-c)
	{
	   x1=x1*10+1;
	}
	    
	    ll x2=pow(10,mn-1);
	    cout<<x2<<" "<<x1*s1;
	    cout<<"\n";
	}

	}
	



return 0; 
}