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 long long h = 31, P = 1e9 + 7;
const long long M = 987654321;
char s[5][2005], t[2005];
bool vis[5][2005];
int n, m, ans;
long long p[2005];
void init() {
  p[0] = 1;
  for (int i = 1; i < 2005; i++) {
    p[i] = p[i - 1] * h % M;
  }
}
struct Hash {
  long long sum[2005];
  void build(char *s) {
    int len = strlen(s + 1);
    sum[0] = 0;
    for (int i = 1; i <= len; i++) {
      sum[i] = (sum[i - 1] * h % M + s[i] - 'a') % M;
    }
  }
  long long get(int l, int r) {
    return (sum[r] - sum[l - 1] * p[r - l + 1] % M + M) % M;
  }
} pre[2], suf[2], hs;
long long f[2][2005][2005];
long long solve(int flag) {
  long long res = 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 <= 1; i++) {
      for (int k = 2; k <= min(n - j + 1, m / 2); k++) {
        if (hs.get(m - k * 2 + 1, m - k) == pre[i].get(j, j + k - 1) &&
            hs.get(m - k + 1, m) ==
                suf[i ^ 1].get(n - (j + k - 1) + 1, n - j + 1)) {
          if (flag || k * 2 != m) res = (res + f[i][j][m - 2 * k]) % P;
        }
      }
    }
    for (int i = 0; i <= 1; i++) {
      for (int k = 2; k <= min(j, m / 2); k++) {
        if (hs.get(k + 1, k * 2) == pre[i].get(j - k + 1, j) &&
            hs.get(1, k) == suf[i ^ 1].get(n - j + 1, n - (j - k + 1) + 1)) {
          if (flag || k * 2 != m) (f[i][j + 1][k * 2] += 1) %= P;
        }
      }
    }
    for (int i = 0; i <= 1; i++) {
      for (int k = 0; k < m; k++) {
        if (s[i][j] == t[k + 1]) {
          (f[i][j + 1][k + 1] += f[i][j][k]) %= P;
          if (k + 2 <= m && s[i ^ 1][j] == t[k + 2]) {
            (f[i ^ 1][j + 1][k + 2] += f[i][j][k]) %= P;
          }
        }
      }
    }
    res = ((res + f[0][j + 1][m]) % P + f[1][j + 1][m]) % P;
  }
  return res;
}
int main() {
  init();
  scanf("%s\n%s\n%s", s[0] + 1, s[1] + 1, t + 1);
  n = strlen(s[0] + 1);
  m = strlen(t + 1);
  hs.build(t);
  for (int i = 0; i <= 1; i++) {
    pre[i].build(s[i]);
    reverse(s[i] + 1, s[i] + n + 1);
    suf[i].build(s[i]);
    reverse(s[i] + 1, s[i] + n + 1);
  }
  long long ans = solve(1);
  if (m > 1) {
    reverse(t + 1, t + m + 1);
    hs.build(t);
    ans = (ans + solve(0)) % P;
    if (m == 2) {
      for (int i = 0; i <= 1; i++) {
        for (int j = 1; j <= n; j++) {
          if (s[i][j] == t[1] && s[i ^ 1][j] == t[2]) {
            ans = (ans - 1 + P) % P;
          }
        }
      }
    }
  }
  cout << ans << endl;
  return 0;
}
