#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
#define f(i,a,b) for( int i=a;i<=b;++i)
#define ff(i,a,b) for( int i=a;i>=b;--i)
#define debug(x) cerr << #x << " : " << x << " " << endl
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<string, string> pss;
const ll mod = 1e9 + 7;
const ll mod2 = 998244353;
const ll inf = 2e18;
const double tiaohe = 0.57721566490153286060651209;
ll oula(ll x) { ll res = x;f(i, 2, x / i) { if (x % i == 0) { res = res / i * (i - 1);while (x % i == 0) x /= i; } }if (x > 1) res = res / x * (x - 1);return res; }
ll quickmod(ll a, ll n, ll m) { ll s = 1;while (n) { if (n & 1) { s = s * a % m; }a = (a*a) % m;n = n / 2; }return s; }
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
void ex_gcd(ll a, ll b, ll &x, ll &y, ll &d) { if (!b) { d = a, x = 1, y = 0; } else { ex_gcd(b, a % b, y, x, d);y -= x * (a / b); } }
ll inv(ll t, ll p) { ll d, x, y;ex_gcd(t, p, x, y, d);return d == 1 ? (x % p + p) % p : -1; }
bool isPrime(ll x) { if (x == 2)return true;if (x % 2 == 0)return false;for (ll i = 2;i*i <= x;i++) if (x % i == 0)return false; return true; }
inline ll in() { char ch = getchar();ll x = 0, f = 1;while (ch<'0' || ch>'9') { if (ch == '-')f = -1;ch = getchar(); }while (ch >= '0'&&ch <= '9') { x = x * 10 + ch - '0';ch = getchar(); }return x * f; }
void print(double x) { printf("%.6lf\n", x); }
//double a = log(n) +tiaohe + 1.0 / (2 * n);
double eqa = (1 + sqrt(5.0)) / 2.0;
double E = 2.7182818284;
const double eps = 1e-8;
double pi = acos(-1);
const int N = 5e5 + 100;
int n, m;
ll k;
int a[N], b[N],id;
ll M[2], LCM;
pair<ll,ll> st[N];//等价类起点
pair<ll,ll> linear( ll B[], ll M[], int n)
{
	//求解 A[i]x = B[i] (mod M[i]), 总共 n 个线性 方程组, → 
	ll x = 0, m = 1;
	for (int i = 0; i < n; i++)
	{
		ll a =  m, b = B[i] -  x, d = gcd(M[i], a);
		if (b % d != 0) return pair<ll,ll>(0, -1);//答案不存在，返回-1 
		ll t = b / d * inv(a / d, M[i] / d) % (M[i] / d);
		x = x + m * t;
		m *= M[i] / d;
	}
	x = (x % m + m) % m;
	return pair<ll, ll>{x,m};//返回的 x 就是答案，m 是最后的 lcm 值
}
ll pos[N][2];
bool check(ll mid) {
	ll res = mid;//减去命中的
	f(i, 1, id) {
		if (st[i].second == -1) {
			continue;
		}
		if (st[i].first > mid)continue;
		ll num = (ll)ceil((mid-1-st[i].first)*1.0/LCM);
		res -= num;
	}
	//cout << mid << " " << res << endl;
	return res <= k;
}
void work(ll k) {
	ll l = 0, r = 2e18,res=0;
	while (l <= r) {
		ll mid = l + r >> 1;
		if (check(mid)) {
			res = mid;
			l = mid + 1;
		}
		else {
			r = mid - 1;
		}
	}
	cout << res << endl;
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
#endif
	cin >> n >> m >> k;
	unordered_map<int, int> mp;
	f(i, 1, n)a[i] = in(), mp[a[i]] = i-1;
	f(i, 1, m)b[i] = in();
	id = 0;
	f(i, 1, m) {
		if (mp.count(b[i])) {
			pos[++id][0] = mp[b[i]];
			pos[id][1] = i-1;
		}
	}
	LCM= 1ll*n * m / gcd(n, m);
	M[0] = n, M[1] = m;
	f(i, 1, id) {
		st[i] = linear(pos[i], M, 2);
		//cout << st[i].first << " " << st[i].second << endl;
	}
	work(k);
	return 0;
}