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:
#include <bits/stdc++.h>
using namespace std;
#define ll long long

ll countdig(ll x){
    ll ret=0;
    while(x>0){
        x/=10;
        ret++;
    }
    return ret;
}

int main() {
	// your code goes here
	int t;
	cin>>t;
	ll mul1=2, mul2=5;
	int pri[9]={3,11,101,1009,11099, 101909, 1120999, 10292809, 113220899};
	while(t--){
	    int a,b,c;
	    cin>>a>>b>>c;
	    a--;
	    b--;
	    float base=pri[c-1];
	    float f=pow(10.0,a)/base;
	    float s=pow(10.0,b)/base;
	    //cout<<f<<' '<<s<<' '<<ceil(log(f)/log(2.0))<<' '<<ceil(log(s)/log(5.0))<<' ';
	    ll fm=ceil(log(f)/log(2.0));
	    ll sm=ceil(log(s)/log(5.0));
	    if(fm<0){
	        fm=0;
	    }
	    if(sm<0){
	        sm=0;
	    }
	    ll an1=base*pow(2,fm);
	    ll an2=base*pow(5,sm);
	    cout<<int(an1)<<' '<<int(an2)<<'\n';
	    
	}
	return 0;
}