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 mod = 1000000007;
inline void add(int &x, int y) { (x += y) >= mod ? x -= mod : 0; }
const unsigned long long sed = 131;
char s1[2020], s2[2020], t[2020];
int n, k, ans;
unsigned long long po[2020];
struct HAS {
  unsigned long long has1[2020], has2[2020];
  void init(char s[]) {
    int len = strlen(s + 1);
    for (int i = 1; i <= len; ++i) has1[i] = has1[i - 1] * sed + s[i];
    for (int i = len; i >= 1; --i) has2[i] = has2[i + 1] * sed + s[i];
  }
  unsigned long long get1(int l, int r) {
    return has1[r] - has1[l - 1] * po[r - l + 1];
  }
  unsigned long long get2(int l, int r) {
    return has2[l] - has2[r + 1] * po[r - l + 1];
  }
} h1, h2, ht;
int dp[2][2020][2][2];
void work() {
  h1.init(s1);
  h2.init(s2);
  ht.init(t);
  int T = 0;
  memset(dp[T], 0, sizeof(dp[T]));
  for (int i = n; i; --i) {
    T ^= 1;
    memset(dp[T], 0, sizeof(dp[T]));
    for (int l = 2; 2 * l <= k && i + l - 1 <= n; ++l) {
      if (h1.get1(i, i + l - 1) * po[l] + h2.get2(i, i + l - 1) ==
          ht.get1(k - 2 * l + 1, k))
        dp[T][k - 2 * l + 1][0][1] = 1;
      if (h2.get1(i, i + l - 1) * po[l] + h1.get2(i, i + l - 1) ==
          ht.get1(k - 2 * l + 1, k))
        dp[T][k - 2 * l + 1][1][1] = 1;
    }
    if (s1[i] == t[k]) dp[T][k][0][0] = 1;
    if (s2[i] == t[k]) dp[T][k][1][0] = 1;
    for (int j = k - 1; j; --j) {
      if (s1[i] == t[j])
        add(dp[T][j][0][0], dp[T ^ 1][j + 1][0][0]),
            add(dp[T][j][0][0], dp[T ^ 1][j + 1][0][1]),
            add(dp[T][j][0][1], dp[T][j + 1][1][0]);
      if (s2[i] == t[j])
        add(dp[T][j][1][0], dp[T ^ 1][j + 1][1][0]),
            add(dp[T][j][1][0], dp[T ^ 1][j + 1][1][1]),
            add(dp[T][j][1][1], dp[T][j + 1][0][0]);
    }
    if (s1[i] == t[1]) {
      add(ans, dp[T][1][0][0]);
      add(ans, dp[T][1][0][1]);
      for (int l = 2; 2 * l < k && i - l >= 0; ++l)
        if (h1.get2(i - l + 1, i) * po[l] + h2.get1(i - l + 1, i) ==
            ht.get1(1, 2 * l))
          add(ans, dp[T][2 * l][1][0]);
    }
    if (s2[i] == t[1]) {
      add(ans, dp[T][1][1][0]);
      add(ans, dp[T][1][1][1]);
      for (int l = 2; 2 * l < k && i - l >= 0; ++l)
        if (h2.get2(i - l + 1, i) * po[l] + h1.get1(i - l + 1, i) ==
            ht.get1(1, 2 * l))
          add(ans, dp[T][2 * l][0][0]);
    }
  }
}
int main() {
  scanf("%s %s %s", s1 + 1, s2 + 1, t + 1);
  n = strlen(s1 + 1);
  k = strlen(t + 1);
  if (k == 1) {
    for (int i = 1; i <= n; ++i) ans += t[1] == s1[i], ans += t[1] == s2[i];
    printf("%d", ans);
    return 0;
  }
  if (k == 2) {
    for (int i = 1; i <= n; ++i) {
      ans += t[1] == s1[i] && t[2] == s1[i + 1];
      ans += t[1] == s1[i] && t[2] == s1[i - 1];
      ans += t[1] == s2[i] && t[2] == s2[i + 1];
      ans += t[1] == s2[i] && t[2] == s2[i - 1];
      ans += t[1] == s1[i] && t[2] == s2[i];
      ans += t[1] == s2[i] && t[2] == s1[i];
    }
    printf("%d", ans);
    return 0;
  }
  po[0] = 1;
  for (int i = 1; i <= 2000; ++i) po[i] = po[i - 1] * sed;
  work();
  reverse(s1 + 1, s1 + 1 + n);
  reverse(s2 + 1, s2 + 1 + n);
  work();
  printf("%d", ans);
  return 0;
}
