Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long N = 2005;
const long long base = 31;
const long long mod = 987654321;
const long long P = 1e9 + 7;
long long n, m, ans, p[N], f[2][N][N];
char s[N], a[2][N];
struct node {
  long long h[N];
  void make(long long n, char *s) {
    h[0] = 0;
    for (long long i = 1; i <= n; i++)
      h[i] = (h[i - 1] * base + s[i] - 'a') % mod;
  }
  long long find(long long l, long long r) {
    return (h[r] - h[l - 1] * p[r - l + 1] % mod + mod) % mod;
  }
} ss, pre[2], nxt[2];
void add(long long &a, long long b) {
  a += b;
  if (a >= P) a -= P;
}
void Pre(long long n) {
  p[0] = 1;
  for (long long i = 1; i <= n; i++) p[i] = p[i - 1] * base % mod;
}
long long solve(long long flag) {
  long long res = 0;
  memset(f, 0, sizeof(f));
  for (long long j = 1; j <= n; j++) {
    f[0][j][0] = f[1][j][0] = 1;
    for (long long i = 0; i < 2; i++)
      for (long long k = 2; k <= min(m / 2, n - j + 1); k++)
        if (ss.find(m - 2 * k + 1, m - k) == pre[i].find(j, j + k - 1) &&
            ss.find(m - k + 1, m) ==
                nxt[i ^ 1].find(n - (j + k - 1) + 1, n - j + 1))
          if (2 * k != m || flag) add(res, f[i][j][m - 2 * k]);
    for (long long i = 0; i < 2; i++)
      for (long long k = 2; k <= min(j, m / 2); k++)
        if (ss.find(1, k) == nxt[i ^ 1].find(n - j + 1, n - (j - k + 1) + 1) &&
            ss.find(k + 1, 2 * k) == pre[i].find(j - k + 1, j))
          if (2 * k != m || flag) add(f[i][j + 1][2 * k], 1);
    for (long long i = 0; i < 2; i++)
      for (long long 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 (long long i = 0; i < 2; i++) add(res, f[i][j + 1][m]);
  }
  return res;
}
signed main() {
  Pre(2000);
  for (long long i = 0; i < 2; i++) {
    cin >> a[i] + 1;
    n = strlen(a[i] + 1);
    pre[i].make(n, a[i]);
    reverse(a[i] + 1, a[i] + n + 1);
    nxt[i].make(n, a[i]);
    reverse(a[i] + 1, a[i] + n + 1);
  }
  cin >> s + 1;
  m = strlen(s + 1);
  ss.make(m, s);
  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 (long long i = 0; i < 2; i++)
        for (long long j = 1; j <= n; j++)
          if (a[i][j] == s[1] && a[i ^ 1][j] == s[2]) add(ans, P - 1);
    }
  }
  cout << ans;
  return 0;
}