// Success consists of going from failure to failure without loss of enthusiasm
#include <bits/stdc++.h>
using namespace std;
#define nl '\n'
using str = string;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int nax = 4105;
using B = bitset<nax>;
using ll = long long;
const ll MX = 1e15+7;
int main() {
cin.tie(0)->sync_with_stdio(0);
int N, M, K; cin >> N >> M >> K;
K = M - K;
auto rnd = [&]() {
return (rng() * 1LL * rng()) % MX;
};
vector<vector<int>> A(N, vector<int>(M));
vector<array<ll, 4>> C(M, {0, 0, 0, 0});
vector<ll> W(N);
for(int i = 0; i < N; i++) {
str S; cin >> S;
W[i] = rnd();
for(int j = 0; j < M; j++) {
int c = -1;
if (S[j] == 'A') c = 0;
if (S[j] == 'G') c = 1;
if (S[j] == 'C') c = 2;
if (S[j] == 'T') c = 3;
A[i][j] = c;
C[j][c] += W[i];
}
}
ll SW = accumulate(begin(W), end(W), 0LL);
for(int i = 0; i < N; i++) {
SW -= W[i];
ll H = 0;
for(int j = 0; j < M; j++) H += C[j][A[i][j]] - W[i];
// cout << SW << " " << H << " " << W[i] << nl;
if (H == (K * SW)) {
cout << i+1 << nl;
return 0;
}
SW += W[i];
}
return 0;
}
