// 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>;
const int MOD = 1e9+7;
int main() {
cin.tie(0)->sync_with_stdio(0);
int N, M, K; cin >> N >> M >> K;
vector<vector<B>> A(N, vector<B>(4));
for(int i = 0; i < N; i++) {
str S; cin >> S;
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][c][j] = 1;
}
}
vector<int> ans(N); iota(begin(ans), end(ans), 0);
auto rnd = [&]() {
return (rng() * 1LL * rng()) % MOD;
};
for(int t = 0; ; t++) {
set<int> in(begin(ans), end(ans));
int n = size(ans);
// cout << n << endl;
int I = ans[rnd() % n];
// cout << I << endl;
int W = 0;
vector<int> nans;
for(int k = 0; k < N; k++) {
int same = 0;
for(int c = 0; c < 4; c++) same += (A[I][c] & A[k][c]).count();
int c = M - same;
if (c == K) {
W++;
if (in.count(k)) nans.push_back(k);
}
}
if (W == N-1) {
cout << I+1 << nl;
return 0;
}
ans.swap(nans);
}
assert(false);
return 0;
}
