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 int                                        long long
#define TEST                                       int t; cin>>t; while(t--)
#define pb                                         push_back
#define pii                                        pair <int, int>
#define vi                                         vector <int>
#define ff                                         first
#define ss                                         second
#define sz(v)                                      (int)v.size()
#define vpii                                       vector <pair <int, int>>
#define vvi                                        vector <vector <int> >
#define all(x)                                     (x).begin(), (x).end()
#define rall(x)                                    (x).rbegin(), (x).rend()
#define endl                                       "\n"
#define MAX                                        1e18
#define MIN                                        INT_MIN
#define mod                                        1000000007
typedef long long                                  ll;



void solve() {

  int a, b, c;
  cin >> a >> b >> c;
  int x = 1, y = 1;
  for (int i = 1; i < a; i++) {
    x *= 10;
  }
  for (int i = 1; i < b; i++) {
    y *= 10;
  }
  int f = 0;
  if (x > y) {
    f = 1;
    swap(x, y);
  }
  y += pow(10, min(a, c) - 1);
  if (f)swap(x, y);
  cout << x << " " << y << endl;
}
signed main()
{
// #ifndef ONLINE_JUDGE
//   freopen("input.txt", "r", stdin);
//   freopen("output.txt", "w", stdout);
// #endif
  mt19937_64 rnd(chrono::system_clock::now().time_since_epoch().count());
  ios_base :: sync_with_stdio(0);
  cin.tie(0);
  //seive();

  TEST
  solve();

}