#include <bits/stdc++.h>
using namespace std;

#define fi first
#define se second
#define mk make_pair

const int MAXN = 5000;
const int INF = 1e9;

int n;

string s[3];

int ok1[MAXN + 5], ok2[MAXN + 5], ok3[MAXN + 5];

int dis[MAXN + 5][MAXN + 5];
const int dx[6] = {1, -1, 0, 0, 1, -1}, dy[6] = {0, 0, 1, -1, 1, -1};

int main() {
	cin >> s[0] >> s[1] >> s[2];
	
	n = s[0].length();
	
	for (int i = 0; i < n; ++i) {
		// consider only s[0] and s[1], is it valid to rotate s[1] i times?
		ok1[i] = true;
		for (int j = 0; j < n; ++j) {
			if (s[0][j] == s[1][(j + i) % n]) {
				ok1[i] = false;
				break;
			}
		}
	}
	
	for (int i = 0; i < n; ++i) {
		// consider only s[0] and s[2], is it valid to rotate s[2] i times?
		ok2[i] = true;
		for (int j = 0; j < n; ++j) {
			if (s[0][j] == s[2][(j + i) % n]) {
				ok2[i] = false;
				break;
			}
		}
	}
	
	for (int i = 0; i < n; ++i) {
		// consider only s[1] and s[2], is it valid to rotate s[2] i times?
		ok3[i] = true;
		for (int j = 0; j < n; ++j) {
			if (s[1][j] == s[2][(j + i) % n]) {
				ok3[i] = false;
				break;
			}
		}
	}
	
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < n; ++j) {
			dis[i][j] = INF;
		}
	}
	
	dis[0][0] = 0;
	
	priority_queue<pair<int, pair<int, int> > > que;
	que.push(mk(0, mk(0, 0)));
	
	while (!que.empty()) {
		pair<int, pair<int, int> > cur = que.top();
		que.pop();
		int x = cur.se.fi;
		int y = cur.se.se;
		if (-cur.fi != dis[x][y]) {
			continue;
		}
		
		for (int i = 0; i < 6; ++i) {
			int xx = (x + dx[i] + n) % n, yy = (y + dy[i] + n) % n;
			if (dis[xx][yy] > dis[x][y] + 1) {
				dis[xx][yy] = dis[x][y] + 1;
				que.push(mk(-dis[xx][yy], mk(xx, yy)));
			}
		}
	}
	
	int ans = INF;
	for (int i = 0; i < n; ++i) { // rotate s[1] i times
		for (int j = 0; j < n; ++j) { // rotate s[2] j times
			int k = ((j >= i) ? j - i : j - i + n); // relative to s[1], s[2] is like rotated k times
			if (ok1[i] && ok2[j] && ok3[k]) {
				// cout << i << " " << j << endl;
				ans = min(ans, dis[i][j]);
			}
		}
	}
	
	if (ans == INF)
		ans = -1;
	cout << ans << endl;
	
	return 0;
}