Oleg Petrov loves crossword puzzles and every Thursday he buys his favorite magazine with crosswords and other word puzzles. In the last magazine Oleg found a curious puzzle, and the magazine promised a valuable prize for it's solution. We give a formal description of the problem below.

The puzzle field consists of two rows, each row contains n cells. Each cell contains exactly one small English letter. You also are given a word w, which consists of k small English letters. A solution of the puzzle is a sequence of field cells c1, ..., ck, such that:

  * For all i from 1 to k the letter written in the cell ci matches the letter wi;
  * All the cells in the sequence are pairwise distinct;
  * For all i from 1 to k - 1 cells ci and ci + 1 have a common side.



Oleg Petrov quickly found a solution for the puzzle. Now he wonders, how many distinct solutions are there for this puzzle. Oleg Petrov doesn't like too large numbers, so calculate the answer modulo 109 + 7.

Two solutions ci and c'i are considered distinct if the sequences of cells do not match in at least one position, that is there is such j in range from 1 to k, such that cj ≠ c'j.

Input

The first two lines contain the state of the field for the puzzle. Each of these non-empty lines contains exactly n small English letters.

The next line is left empty.

The next line is non-empty and contains word w, consisting of small English letters.

The length of each line doesn't exceed 2 000.

Output

Print a single integer — the number of distinct solutions for the puzzle modulo 109 + 7.

Examples

Input

code
edoc

code


Output

4


Input

aaa
aaa

aa


Output

14
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2111;
const int mod = 1e9 + 7;
const int mod2 = 998244353;
const int base = 443;
inline void add(int& x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
namespace HS {
int pw[maxn], pw2[maxn];
void prework() {
  pw[0] = pw2[0] = 1;
  for (int i = 1; i < maxn; i++) {
    pw[i] = (long long)pw[i - 1] * base % mod;
    pw2[i] = (long long)pw2[i - 1] * base % mod2;
  }
}
inline pair<int, int> calhs(pair<int, int>* hs, int l, int r) {
  return make_pair(
      ((hs[r].first - (long long)hs[l - 1].first * pw[r - l + 1]) % mod + mod) %
          mod,
      ((hs[r].second - (long long)hs[l - 1].second * pw2[r - l + 1] % mod2 +
        mod2) %
       mod2));
}
void construct(char* s, pair<int, int>* hs, int n) {
  hs[0] = make_pair(0, 0);
  for (int i = 1; i <= n; i++) {
    hs[i] =
        make_pair(((long long)hs[i - 1].first * base + s[i] - 'a' + 1) % mod,
                  ((long long)hs[i - 1].second * base + s[i] - 'a' + 1) % mod2);
  }
}
}  // namespace HS
char s[2][maxn], rs[2][maxn], t[maxn];
int n, m;
pair<int, int> hs[2][maxn], rhs[2][maxn], hst[maxn];
bool cal_front(int i, int j, int l) {
  if (l % 2) return false;
  l /= 2;
  if (j < l) return false;
  return (HS::calhs(hs[i], j - l + 1, j) == HS::calhs(hst, l + 1, l + l) &&
          HS::calhs(rhs[i ^ 1], n - j + 1, n - j + l) == HS::calhs(hst, 1, l));
}
bool cal_back(int i, int j, int l) {
  if (l % 2) return false;
  l /= 2;
  if (l == 0) return true;
  if (j + l > n + 1) return false;
  return (HS::calhs(hs[i], j, j + l - 1) ==
              HS::calhs(hst, m - l - l + 1, m - l) &&
          HS::calhs(rhs[i ^ 1], n - j - l + 2, n - j + 1) ==
              HS::calhs(hst, m - l + 1, m));
}
int dp[2][maxn][maxn], ans;
void calc() {
  memset(dp, 0, sizeof(dp));
  for (int j = 1; j <= n; j++)
    for (int i = 0; i < 2; i++) {
      dp[i][j][1] = 1;
      for (int k = 5; k <= m; k += 2)
        if (cal_front(i, j - 1, k - 1)) {
          add(dp[i][j][k], 1);
        }
      for (int k = 1; k <= m; k++)
        if (s[i][j] == t[k]) {
          add(dp[i][j + 1][k + 1], dp[i][j][k]);
          if (s[i ^ 1][j] == t[k + 1])
            add(dp[i ^ 1][j + 1][k + 2], dp[i][j][k]);
        }
    }
  for (int i = 0; i < 2; i++)
    for (int j = 1; j <= n + 1; j++)
      for (int k = 1; k <= m + 1; k++)
        if (cal_back(i, j, m - k + 1) && m - k + 1 != 2) add(ans, dp[i][j][k]);
}
int main() {
  scanf("%s%s%s", s[0] + 1, s[1] + 1, t + 1);
  n = strlen(s[0] + 1);
  m = strlen(t + 1);
  if (m == 1) {
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      if (s[0][i] == t[1]) ans++;
      if (s[1][i] == t[1]) ans++;
    }
    printf("%d\n", ans);
    return 0;
  }
  if (m == 2) {
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      if (s[0][i] == t[1] && s[0][i + 1] == t[2]) ans++;
      if (s[0][i] == t[1] && s[0][i - 1] == t[2]) ans++;
      if (s[1][i] == t[1] && s[1][i + 1] == t[2]) ans++;
      if (s[1][i] == t[1] && s[1][i - 1] == t[2]) ans++;
      if (s[0][i] == t[1] && s[1][i] == t[2]) ans++;
      if (s[1][i] == t[1] && s[0][i] == t[2]) ans++;
    }
    printf("%d\n", ans);
    return 0;
  }
  memcpy(rs, s, sizeof(s));
  reverse(rs[0] + 1, rs[0] + n + 1);
  reverse(rs[1] + 1, rs[1] + n + 1);
  HS::prework();
  HS::construct(s[0], hs[0], n);
  HS::construct(s[1], hs[1], n);
  HS::construct(rs[0], rhs[0], n);
  HS::construct(rs[1], rhs[1], n);
  HS::construct(t, hst, m);
  calc();
  swap(s[0], rs[0]);
  swap(s[1], rs[1]);
  swap(hs[0], rhs[0]);
  swap(hs[1], rhs[1]);
  calc();
  printf("%d\n", ans);
  return 0;
}
