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
#define ld long double
#define pb push_back
#define vll vector<ll>
#define pll pair<ll,ll>
#define vpll vector<pll>
#define ub upper_bound
#define lb lower_bound
#define all(v) ((v).begin()),((v).end())
#define allr(v) ((v).rbegin()),((v).rend())
#define ff first
#define ss second
#define mp make_pair
#define pll pair<ll,ll> 
#define fo(i,n) for(ll i=0;i<n;i++)
#define foa(i,s,e) for(ll i=(s);i<=e;i++)
#define fod(i,s,e) for(ll i= (s);i>=(e);i--)
#define max3(a,b,c) max(max(a,b),c)
#define min3(a,b,c) min(min(a,b),c)
#define deb(x) cerr<<#x<<' '<<'='<<' '<<x<<'\n'
#define sz(x) (ll)(x.size())
#define ANS cout<<ans<<'\n'
#define YES cout<<"YES\n"
#define NO cout<<"NO\n"
#define Yes cout<<"Yes\n"
#define No cout<<"No\n"
const ld pi = 3.14159265358979323846;
ll MOD = 1e9 + 7;
//ll MOD = 998244353;
const char nl = '\n';
const ll inf = 2000000000000000000;
#define fill(a,b) memset(a, b, sizeof(a))
#define setbits(x) __builtin_popcountll(x)
#define vout(v) for(auto &x : v){cout << x << " ";}
#define vin(v) for (auto &x : v){cin >> x;}
#define print2d(dp,n,m) fo(i, n){fo(j, m)cout<<dp[i][j]<<" ";cout<<"\n";}
ll nxt(){ll x; cin >> x; return x;}
ll mul(ll x,ll y){
    return (1ll* (x%MOD)*(y%MOD));
}
ll modpow(ll x,ll y){ll z=1;while(y>0){if(y%2)z=mul(z,x);x =mul(x,x) ;y/=2;}return z;}
ll power(ll x,ll y){ll z=1;while(y>0){if(y%2)z=z*x;x =x*x ;y/=2;}return z;}
ll gcd(ll a,ll b){if(a<b) return gcd(b,a);if(b==0) return a;return gcd(b,a%b);} 
ll min(ll a,ll b){if(a<b)return a;return b;}
ll max(ll a,ll b){if(a>b)return a;return b;}
ll sq(ll a){
    ll ans = (1ll*a*a);
    return ans;
}
void solve(){
    ll a, b, c;
    cin >> a >> b >> c;
        string sa = "1";
        fo(i, a-1)sa+="0";
        string sb = "1";
        fo(i, b-1)sb+="0";
        string sc = "1";
        fo(i, c-1)sc+="0";
        if(a >= b)sb[b-c] = '1';
        else sa[a-c] = '1';
        cout << sa << " " << sb << nl;    
}
int main(){ 
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    #ifndef ONLINE_JUDGE
    freopen("inputf.txt" , "r" , stdin) ;
    freopen("outputf.txt" , "w" , stdout) ;
    freopen("error.txt" , "w" , stderr) ;
    #endif
    ll TC = 1;
    cin>>TC;
    fo(TT, TC){
        //cout << "Case #"<< TT + 1 << ": ";
        solve();
    }
    cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}