Problem: In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.

You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.

Input

The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.

Output

Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted. 

If there is no valid common subsequence, output 0.

Examples

Input

AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ


Output

ORZ


Input

AA
A
A


Output

0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("no-stack-protector")
#pragma GCC target("sse,sse2,ssse3,sse3,sse4,popcnt,avx,mmx,abm,tune=native")
using namespace std;
namespace output {
void __(short x) { cout << x; }
void __(unsigned x) { cout << x; }
void __(int x) { cout << x; }
void __(long long x) { cout << x; }
void __(unsigned long long x) { cout << x; }
void __(double x) { cout << x; }
void __(long double x) { cout << x; }
void __(char x) { cout << x; }
void __(const char* x) { cout << x; }
void __(const string& x) { cout << x; }
void __(bool x) { cout << (x ? "true" : "false"); }
template <class S, class T>
void __(const pair<S, T>& x) {
  __(1 ? "(" : ""), __(x.first), __(1 ? ", " : " "), __(x.second),
      __(1 ? ")" : "");
}
template <class T>
void __(const vector<T>& x) {
  __(1 ? "{" : "");
  bool _ = 0;
  for (const auto& v : x) __(_ ? 1 ? ", " : " " : ""), __(v), _ = 1;
  __(1 ? "}" : "");
}
template <class T>
void __(const set<T>& x) {
  __(1 ? "{" : "");
  bool _ = 0;
  for (const auto& v : x) __(_ ? 1 ? ", " : " " : ""), __(v), _ = 1;
  __(1 ? "}" : "");
}
template <class T>
void __(const multiset<T>& x) {
  __(1 ? "{" : "");
  bool _ = 0;
  for (const auto& v : x) __(_ ? 1 ? ", " : " " : ""), __(v), _ = 1;
  __(1 ? "}" : "");
}
template <class S, class T>
void __(const map<S, T>& x) {
  __(1 ? "{" : "");
  bool _ = 0;
  for (const auto& v : x) __(_ ? 1 ? ", " : " " : ""), __(v), _ = 1;
  __(1 ? "}" : "");
}
void pr() { cout << "\n"; }
template <class S, class... T>
void pr(const S& a, const T&... b) {
  __(a);
  if (sizeof...(b)) __(' ');
  pr(b...);
}
}  // namespace output
using namespace output;
const int MN = 105;
int dp[MN][MN][MN], i, j, k, n, m, l, ans, f[MN];
struct idk {
  int x, y, w;
} ptr[MN][MN][MN], sna;
string s, t, v;
int main() {
  cin >> s >> t >> v;
  n = s.size(), m = t.size(), l = v.size();
  s.insert(s.begin(), ' ');
  t.insert(t.begin(), ' ');
  f[0] = 0;
  for (i = 1, j = 0; i < l; i++) {
    while (j > 0 && v[i] != v[j]) j = f[j - 1];
    if (j >= 0 && v[i] == v[j])
      f[i] = ++j;
    else
      f[i] = j++;
  }
  for (i = 0; i <= n; i++) {
    for (j = 0; j <= m; j++) {
      for (k = 1; k <= l; k++) dp[i][j][k] = -1 << 30;
      if (i && j) dp[i][j][0] = -1 << 30;
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      for (k = 0; k <= l; k++) {
        if (dp[i - 1][j][k] > dp[i][j][k]) {
          dp[i][j][k] = dp[i - 1][j][k];
          ptr[i][j][k] = {i - 1, j, k};
        }
        if (dp[i][j - 1][k] > dp[i][j][k]) {
          dp[i][j][k] = dp[i][j - 1][k];
          ptr[i][j][k] = {i, j - 1, k};
        }
        if (s[i] == t[j]) {
          if (k == l) continue;
          int t = k;
          while (s[i] != v[t] && t > 0) t = f[t - 1];
          if (s[i] == v[t]) {
            if (dp[i - 1][j - 1][k] + 1 > dp[i][j][t + 1]) {
              dp[i][j][t + 1] = dp[i - 1][j - 1][k] + 1;
              ptr[i][j][t + 1] = {i - 1, j - 1, k};
            }
          } else {
            if (dp[i - 1][j - 1][k] + 1 > dp[i][j][0]) {
              dp[i][j][0] = dp[i - 1][j - 1][k] + 1;
              ptr[i][j][0] = {i - 1, j - 1, k};
            }
          }
        }
      }
      for (k = 0; k < l; k++) {
        if (dp[i][j][k] > ans) {
          ans = dp[i][j][k];
          sna = {i, j, k};
        }
      }
    }
  }
  if (ans == 0)
    printf("0\n");
  else {
    int x = sna.x, y = sna.y, w = sna.w;
    string res;
    while (x && y) {
      int px = ptr[x][y][w].x;
      int py = ptr[x][y][w].y;
      int pw = ptr[x][y][w].w;
      if (dp[px][py][pw] < dp[x][y][w]) res += s[x];
      x = px, y = py, w = pw;
    }
    reverse(res.begin(), res.end());
    printf("%s\n", res.c_str());
  }
  return 0;
}