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 N = 2123;
const int P = 1000000007;
char s[2][N], t[N];
int n, m;
int lcp[2][N][N], lcs[2][N][N];
int f[N][N][2];
void upd(int &x, int y) {
  x += y;
  if (x >= P) x -= P;
}
int work(bool recount) {
  for (int k = 0; k < 2; ++k)
    for (int i = n - 1; i >= 0; --i)
      for (int j = m - 1; j >= 0; --j)
        if (s[k][i] == t[j])
          lcp[k][i][j] = lcp[k][i + 1][j + 1] + 1;
        else
          lcp[k][i][j] = 0;
  reverse(t, t + m);
  for (int k = 0; k < 2; ++k)
    for (int i = n - 1; i >= 0; --i)
      for (int j = m - 1; j >= 0; --j)
        if (s[k][i] == t[j])
          lcs[k][i][j] = lcs[k][i + 1][j + 1] + 1;
        else
          lcs[k][i][j] = 0;
  reverse(t, t + m);
  memset(f, 0, sizeof(f));
  int ans = 0;
  for (int i = 0; i < n + 1; ++i) {
    for (int k = 0; k < 2; ++k) f[i][0][k] = 1;
    for (int j = 2; j < i + 1; ++j)
      if (!recount || j * 2 < m)
        for (int k = 0; k < 2; ++k) {
          if (lcp[k][i - j][j] >= j && lcs[!k][i - j][m - j] >= j)
            upd(f[i][2 * j][k], 1);
        }
    for (int j = 0; j < m; ++j)
      for (int k = 0; k < 2; ++k)
        if (i < n && f[i][j][k] > 0 && s[k][i] == t[j]) {
          upd(f[i + 1][j + 1][k], f[i][j][k]);
          if (j + 2 <= m && s[!k][i] == t[j + 1])
            upd(f[i + 1][j + 2][!k], f[i][j][k]);
          if (m - j > 2 && (m - j) % 2 == 0 && (!recount || j > 0)) {
            int l = (m - j) / 2;
            if (lcp[k][i][j] >= l && lcs[!k][i][0] >= l) upd(ans, f[i][j][k]);
          }
        }
    for (int k = 0; k < 2; ++k) upd(ans, f[i][m][k]);
  }
  return ans;
}
int main() {
  for (int i = 0; i < 2; ++i) scanf("%s", s[i]);
  scanf("%s", t);
  n = strlen(s[0]), m = strlen(t);
  int ans = work(false);
  if (m > 1) {
    reverse(t, t + m);
    ans = (ans + work(true)) % P;
  }
  if (m == 2) {
    for (int i = 0; i < n; ++i) {
      if (s[0][i] == t[0] && s[1][i] == t[1]) --ans;
      if (s[0][i] == t[1] && s[1][i] == t[0]) --ans;
    }
  }
  printf("%d\n", (ans + P) % P);
  return 0;
}
