#include <bits/stdc++.h>
using namespace std;
#define bug(x) cerr<<#x<<" : "<<x<<endl;
#define PYES puts("YES");
#define pyes puts("yes");
#define PNO puts("YES");
#define pno puts("no");
const int N=2e6+10; 
const int M=1e4+5;
const int mod=998244353;
typedef long long ll;

//ll a[N],ans[N];
//int a[N][N];
//char g[N][N];
int a[N];
//求 m^k mod p，时间复杂度 O(logk)。

ll qpow(ll a, ll b) {
    ll ans = 1%mod;
    while (b) {
        if (b & 1) ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}
ll getInv(ll a) { return qpow(a, mod - 2); }  //求一个数的逆元

int main(){
    //freopen("in.txt", "r", stdin);
     //freopen("std.txt", "w", stdout);
    int T;
    cin>>T;
    
    while(T--){
        ll aa,bb,k;
        cin>>aa>>bb>>k;
        if(aa==0&&bb==0){
            cout<<0<<" "<<0<<endl;
            continue;
        }
        if(k&1){
            ll aaa=aa;
            aa=(qpow(2,k/2)*aa%mod+qpow(2,k/2)*bb%mod)%mod;
            bb=(qpow(2,k/2)*aaa%mod+mod-qpow(2,k/2)*bb%mod)%mod;
        }
        else{
            ll aaa=aa;
            aa=(qpow(2,k/2)*aa%mod)%mod;
            bb=(qpow(2,k/2)*bb%mod)%mod;
        }
        cout<<aa<<" "<<bb<<endl;
    }    
}

/*





*/