#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 10;
string A, B;                       //原始串
string a[N], b[N];                 //规则
queue<string> qa, qb;              //双端队列
unordered_map<string, int> da, db; //此字符串，是几步转移过来的
int n;
int ans = INF;

inline bool expandA() {
    string u = qa.front();
    qa.pop();
    if (db.count(u)) {
        ans = da[u] + db[u];
        return true;
    }
    for (int i = 0; i < u.size(); i++) //枚举字符串的每一位
        for (int j = 0; j < n; j++) {  //枚举规则
            if (u.substr(i, a[j].size()) == a[j]) {
                string ts = u.substr(0, i) + b[j] + u.substr(i + a[j].size());
                if (!da.count(ts)) {
                    qa.push(ts);
                    da[ts] = da[u] + 1;
                }
            }
        }
    return false;
}

inline bool expandB() {
    string u = qb.front();
    qb.pop();
    if (da.count(u)) {
        ans = da[u] + db[u];
        return true;
    }
    for (int i = 0; i < u.size(); i++)
        for (int j = 0; j < n; j++) {
            if (u.substr(i, b[j].size()) == b[j]) {
                string ts = u.substr(0, i) + a[j] + u.substr(i + b[j].size());
                if (!db.count(ts)) {
                    qb.push(ts);
                    db[ts] = db[u] + 1;
                }
            }
        }
    return false;
}

void bfs() {
    //两个串分别入队列
    qa.push(A), qb.push(B);
    da[A] = 0, db[B] = 0;

    //双向广搜套路
    while (qa.size() && qb.size()) {
        if (qa.size() < qb.size()) {
            if (expandA()) return;
            if (expandB()) return;
        } else {
            if (expandB()) return;
            if (expandA()) return;
        }
    }
}

//可以AC掉本题,双向广搜+大小队列对比优化
// 17ms
int main() {
    //加快读入
    cin.tie(0), ios::sync_with_stdio(false);

    cin >> A >> B;
    while (cin >> a[n] >> b[n]) n++;

    bfs();

    if (ans > 10)
        puts("NO ANSWER!");
    else
        printf("%d\n", ans);
    return 0;
}