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:
#pragma region templates
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
template<typename T> using V = vector<T>;
template<typename T> using VV = vector<vector<T>>;
template<typename T> using PQ = priority_queue<T>;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define lb(c, x) distance((c).begin(), lower_bound(all(c), (x)))
#define ub(c, x) distance((c).begin(), upper_bound(all(c), (x)))
#define all(v) (v).begin(),(v).end()
#define siz(v) (ll)(v).size()
#define rep(i,a,n) for(ll i=a;i<(ll)(n);++i)
#define repr(i,a,n) for(ll i=n-1;(ll)a<=i;--i)
#define ENDL '\n'
typedef pair<int,int> Pi;
typedef pair<ll,ll> PL;
constexpr ll mod = 1000000007; // 998244353;
constexpr ll INF = 1000000099;
constexpr ll LINF = (ll)(1e18 +99);
const ld PI = acos((ld)-1);
constexpr ll dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};
template<typename T,typename U> inline bool chmin(T& t, const U& u){if(t>u){t=u;return 1;}return 0;}
template<typename T,typename U> inline bool chmax(T& t, const U& u){if(t<u){t=u;return 1;}return 0;}
template<typename T> inline T gcd(T a,T b){return b?gcd(b,a%b):a;}
inline void Yes() { cout << "Yes" << ENDL; }
inline void No() { cout << "No" << ENDL; }
inline void YES() { cout << "YES" << ENDL; }
inline void NO() { cout << "NO" << ENDL; }

template<typename T,typename Y> inline T mpow(T a, Y n) {
  T res = 1;
  for(;n;n>>=1) {
    if (n & 1) res = res * a;
    a = a * a;
  }
  return res;
}

template <typename T>
vector<T> finddivisor(T x) {  //整数xの約数(xを含む)
  vector<T> divisor;
  for(T i = 1; (i * i) <= x; i++) {
    if(x % i == 0) {
      divisor.push_back(i);
      if(i * i != x) { divisor.push_back(x / i);}
    }
  }
  sort(divisor.begin(), divisor.end());
  return divisor;
}

template <typename T> V<T> prefix_sum(const V<T>& v) {
  int n = v.size();
  V<T> ret(n + 1);
  rep(i, 0, n) ret[i + 1] = ret[i] + v[i];
  return ret;
}


template<typename T>
T rand(T l,T r){
  static random_device rd;
  static mt19937 g(rd());
  return uniform_int_distribution<T>(l,r)(g);
}

template<typename T>
istream& operator >> (istream& is, vector<T>& vec){
  for(auto&& x: vec) is >> x;
  return is;
}

template<typename T,typename Y>
ostream& operator<<(ostream& os,const pair<T,Y>& p){
  return os<<"{"<<p.fs<<","<<p.sc<<"}";
}

template<typename T> ostream& operator<<(ostream& os,const V<T>& v){
  os<<"{";
  for(auto e:v)os<<e<<",";
  return os<<"}";
}

template<typename ...Args>
void debug(Args&... args){
  for(auto const& x:{args...}){
    cerr<<x<<' ';
  }
  cerr<<ENDL;
}
#pragma endregion templates

signed main(){
  cin.tie(0);cerr.tie(0);ios::sync_with_stdio(false);
  cout<<fixed<<setprecision(20);
  
  int qq;cin>>qq;
  while(qq--){
    ll a,b,c;cin>>a>>b>>c;

    if(min(a,b)==c){
      cout<<mpow(10ll,a-1)<<" "<<mpow(10ll,b-1)<<ENDL;
    }else{
      ll x=mpow(10ll,a-1),y=mpow(10ll,b-1);
      y+=mpow(10ll,c-1);

      cout<<x<<" "<<y<<ENDL;
    }
  }
}
//(・_・)(・_・)(・_・)(・_・)
//CHECK overflow,what to output?
//any other simpler approach?