一開始用bitset水了50分。。。但想一想那個年代stl還沒有解禁啊。。。
然後不會，棄療，看題解，竟然是矩陣！
生成行向量n^2判斷矩陣相等太妙了
判斷時哪裡不相等就把它拎出來暴力判
k=3時平方就與k=2時基本相同了
自己寫..wa…然後抄程式碼….

#include<bits/stdc++.h>
#define rep(i,n) for(int i=1;i<=(n);i++)
using namespace std;
const int N=200005;
void sc(int& x){
    x=0;char c=getchar();
    for(;c>'9'
 
||c<'0';c=getchar());
    for(;c<='9' && c>='0';c=getchar())x=10*x+c-'0';
}
int n,d,k,C[N],B[N],AA[N],A[N][110],lim,cnt=0,id[110][110];
void upd(int& x,int y){x+=y;if(x>=k)x-=k;while(x<0)x+=k;}
void solve2(){
    rep(i,n)rep(j,d)upd(B[j],A[i][j]);
    rep(i,n)rep(j,d)upd(C[i],B[j]*A[i][j]);
    rep(i,n)upd(C[i],-AA[i]);lim=(n-1
 
)%k;
}
void solve3(){
    rep(i,d)rep(j,d)id[i][j]=++cnt;
    rep(i,n)rep(j,d)rep(p,d)upd(B[id[j][p]],A[i][j]*A[i][p]);
    rep(i,n)rep(j,d)rep(p,d)upd(C[i],B[id[j][p]]*A[i][j]*A[i][p]);
    rep(i,n)upd(C[i],-AA[i]*AA[i]);lim=(n-1)%k;
}
bitset<205> a[N],tmp;
void solve4(){
    rep(i,n){int x;
    rep(j,d){
        sc(x);a[i][j]=x%k;
    }
    rep(j,i-1
 
){
        tmp=a[i]&a[j];
        if(tmp.count()%k==0){
        printf("%d %d",j,i);return;
        }
    }
    }
    printf("-1 -1\n");
}
int main(){//freopen("in.in","r",stdin);
    sc(n),sc(d),sc(k);
    if(n<=1000 && k==2){solve4();return 0;}
    rep(i,n)rep(j,d)sc(A[i][j]),A[i][j]%=k,AA[i]+=A[i][j]*A[i][j]%k;
    if(k==2)solve2();else solve3();
    rep(i,n)if(C[i]!=lim){
        rep(j,n)if(j!=i){
        int ans=0;
        rep(p,d)ans+=A[i][p]*A[j][p];
        if(ans%k==0){if(j<i)printf("%d %d\n",j,i);else printf("%d %d\n",i,j);return 0;}
    }
    }
    printf("-1 -1\n");
}