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;
inline char nc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read(char *s) {
  char c = nc();
  int len = 0;
  for (; !(c >= 'a' && c <= 'z'); c = nc())
    ;
  for (; c >= 'a' && c <= 'z'; s[++len] = c, c = nc())
    ;
  s[++len] = 0;
  return len - 1;
}
const int P = 1e9 + 7;
const int N = 2005;
inline void add(int &x, int y) {
  x += y;
  if (x >= P) x -= P;
}
const long long S = 31;
const long long MOD = 987654321;
long long seed[N];
inline void Pre(int n) {
  seed[0] = 1;
  for (int i = (1); i <= (n); ++i) seed[i] = seed[i - 1] * S % MOD;
}
struct Hash {
  long long h[N];
  inline void make(int n, char *s) {
    h[0] = 0;
    for (int i = (1); i <= (n); ++i) h[i] = (h[i - 1] * S + s[i] - 'a') % MOD;
  }
  inline long long cut(int l, int r) {
    return (h[r] + MOD - h[l - 1] * seed[r - l + 1] % MOD) % MOD;
  }
} pre[2], suf[2], ss;
int n, m;
char a[2][N], s[N];
int f[2][N][N];
inline int Solve(int flag) {
  int ret = 0;
  memset(f, 0, sizeof(f));
  for (int j = (1); j <= (n); ++j) {
    f[0][j][0] = f[1][j][0] = 1;
    for (int i = (0); i < (2); ++i)
      for (int k = (2); k <= (min(n - j + 1, m / 2)); ++k)
        if (ss.cut(m - 2 * k + 1, m - k) == pre[i].cut(j, j + k - 1) &&
            ss.cut(m - k + 1, m) ==
                suf[i ^ 1].cut(n - (j + k - 1) + 1, n - j + 1))
          if (2 * k != m || flag) add(ret, f[i][j][m - 2 * k]);
    for (int i = (0); i < (2); ++i)
      for (int k = (2); k <= (min(j, m / 2)); ++k)
        if (ss.cut(k + 1, 2 * k) == pre[i].cut(j - k + 1, j) &&
            ss.cut(1, k) == suf[i ^ 1].cut(n - j + 1, n - (j - k + 1) + 1))
          if (2 * k != m || flag) add(f[i][j + 1][2 * k], 1);
    for (int i = (0); i < (2); ++i)
      for (int k = (0); k < (m); ++k)
        if (a[i][j] == s[k + 1]) {
          add(f[i][j + 1][k + 1], f[i][j][k]);
          if (k + 2 <= m && a[i ^ 1][j] == s[k + 2])
            add(f[i ^ 1][j + 1][k + 2], f[i][j][k]);
        }
    for (int i = (0); i < (2); ++i) add(ret, f[i][j + 1][m]);
  }
  return ret;
}
int main() {
  Pre(2000);
  n = read(a[0]);
  read(a[1]);
  for (int i = (0); i < (2); ++i) {
    pre[i].make(n, a[i]);
    reverse(a[i] + 1, a[i] + n + 1);
    suf[i].make(n, a[i]);
    reverse(a[i] + 1, a[i] + n + 1);
  }
  m = read(s);
  ss.make(m, s);
  int Ans = 0;
  add(Ans, Solve(1));
  if (m > 1) {
    reverse(s + 1, s + m + 1);
    ss.make(m, s);
    add(Ans, Solve(0));
    if (m == 2) {
      for (int j = (1); j <= (n); ++j)
        for (int i = (0); i < (2); ++i)
          if (a[i][j] == s[1] && a[i ^ 1][j] == s[2]) add(Ans, P - 1);
    }
  }
  printf("%d\n", Ans);
  return 0;
}
