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 = 1e9 + 7;
const int base = 233;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
int n, m, pre[2][2020], suf[2][2020];
int h[2020], c[2020];
char s[2][2020], t[2020];
int Solve_1() {
  int ans = 0;
  for (int i = 0; i <= 1; ++i)
    for (int j = 1; j <= n; ++j) ans += s[i][j] == t[1];
  return ans;
}
int Solve_2() {
  int ans = 0;
  for (int i = 0; i <= 1; ++i) {
    for (int j = 1; j <= n; ++j) {
      ans += s[i][j] == t[1] && s[i ^ 1][j] == t[2];
      ans += s[i][j] == t[1] && s[i][j + 1] == t[2];
      ans += s[i][j] == t[1] && s[i][j - 1] == t[2];
    }
  }
  return ans;
}
int splice(int h1, int h2, int x) { return (h2 + 1LL * h1 * c[x]) % mod; }
int Get1(int l, int r, int i) {
  return (pre[i][r] - 1LL * pre[i][l - 1] * c[r - l + 1] % mod + mod) % mod;
}
int Get2(int l, int r, int i) {
  return (suf[i][l] - 1LL * suf[i][r + 1] * c[r - l + 1] % mod + mod) % mod;
}
int qwq, f[2][2020][2020], g[2][2020][2020];
int Solve() {
  memset(f, 0, sizeof(f));
  memset(g, 0, sizeof(g));
  for (int i = 1; i <= m; ++i) {
    h[i] = (1LL * h[i - 1] * base + (t[i] - 'a' + 1)) % mod;
  }
  for (int i = 0; i <= 1; ++i) {
    for (int j = 1; j <= n; ++j) {
      for (int k = j; k >= 1; --k) {
        int len = 2 * (j - k + 1);
        if (len > m) continue;
        int Ha = Get1(k, j, i);
        int Hb = Get2(k, j, i ^ 1);
        int Ht = h[len];
        if (splice(Hb, Ha, j - k + 1) == Ht) {
          if (len == m) {
            ++qwq;
          } else {
            g[i][j][len] = 1;
          }
        }
      }
    }
  }
  for (int i = 0; i <= 1; ++i) {
    for (int j = 1; j <= n; ++j) {
      g[i][j][1] = (s[i][j] == t[1]);
    }
  }
  for (int k = 1; k < m; ++k) {
    for (int i = 0; i <= 1; ++i) {
      for (int j = 1; j <= n; ++j) {
        if (f[i][j][k]) {
          if (s[i][j + 1] == t[k + 1])
            f[i][j + 1][k + 1] = (f[i][j + 1][k + 1] + f[i][j][k]) % mod;
          if (s[i ^ 1][j] == t[k + 1])
            g[i ^ 1][j][k + 1] = (g[i ^ 1][j][k + 1] + f[i][j][k]) % mod;
        }
        if (g[i][j][k]) {
          if (s[i][j + 1] == t[k + 1])
            f[i][j + 1][k + 1] = (f[i][j + 1][k + 1] + g[i][j][k]) % mod;
        }
      }
    }
  }
  for (int i = m; i >= 1; --i) {
    h[i] = (1LL * h[i + 1] * base % mod + (t[i] - 'a' + 1)) % mod;
  }
  int ans = 0;
  for (int i = 0; i <= 1; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (s[i][j] == t[m]) {
        ans = (ans + (f[i][j - 1][m - 1] + g[i][j - 1][m - 1]) % mod) % mod;
      }
      for (int k = j; k <= n; ++k) {
        int len = 2 * (k - j + 1) % mod;
        if (len > m) continue;
        int Ha = Get1(j, k, i ^ 1);
        int Hb = Get2(j, k, i);
        int Ht = h[m - len + 1];
        if (splice(Ha, Hb, k - j + 1) == Ht) {
          if (len == m) {
            ++qwq;
          } else {
            ans = (ans + (f[i][j - 1][m - len] + g[i][j - 1][m - len]) % mod) %
                  mod;
          }
        }
      }
    }
  }
  return ans;
}
int main() {
  scanf("%s", s[0] + 1);
  scanf("%s", s[1] + 1);
  scanf("%s", t + 1);
  n = strlen(s[0] + 1);
  m = strlen(t + 1);
  if (m == 1) return !printf("%d\n", Solve_1());
  if (m == 2) return !printf("%d\n", Solve_2());
  c[0] = 1;
  for (int i = 1; i <= n; ++i) {
    c[i] = 1LL * c[i - 1] * base % mod;
  }
  for (int i = 0; i <= 1; ++i) {
    for (int j = 1; j <= n; ++j) {
      pre[i][j] = (1LL * pre[i][j - 1] * base + (s[i][j] - 'a' + 1)) % mod;
    }
    for (int j = n; j >= 1; --j) {
      suf[i][j] = (1LL * suf[i][j + 1] * base + (s[i][j] - 'a' + 1)) % mod;
    }
  }
  int ans = Solve();
  reverse(t + 1, t + m + 1);
  ans = (ans + Solve()) % mod;
  ans = (ans + (qwq >> 1)) % mod;
  printf("%d\n", ans);
  return 0;
}
