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;
template <typename _T>
inline void read(_T &f) {
  f = 0;
  _T fu = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') {
      fu = -1;
    }
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    f = (f << 3) + (f << 1) + (c & 15);
    c = getchar();
  }
  f *= fu;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + 48);
  else
    print(x / 10), putchar(x % 10 + 48);
}
template <typename T>
void print(T x, char t) {
  print(x);
  putchar(t);
}
const int N = 2005, md = 1e9 + 7, base = 19491001, inv2 = (md + 1) >> 1;
inline int add(int x, int y) {
  x += y;
  if (x >= md) x -= md;
  return x;
}
inline void addx(int &x, int y) {
  x += y;
  if (x >= md) x -= md;
}
inline int sub(int x, int y) {
  x -= y;
  if (x < 0) x += md;
  return x;
}
inline int mul(int x, int y) { return 1ll * x * y % md; }
inline int fpow(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = mul(ans, x);
    y >>= 1;
    x = mul(x, x);
  }
  return ans;
}
struct hash_t {
  char c[N];
  unsigned long long ha[N], pw[N];
  int n, opt;
  void ini(int a, int b) {
    n = a;
    opt = b;
    pw[0] = 1;
    for (int i = 1; i <= n; i++) {
      pw[i] = pw[i - 1] * base;
      ha[i] = ha[i - 1] * base + c[i] + 233;
    }
  }
  inline unsigned long long qry(int l, int r) {
    if (opt) l = n - l + 1, r = n - r + 1;
    return ha[r] - ha[l - 1] * pw[r - l + 1];
  }
} ha[2][2], he;
int f[2][N][2];
char c[2][N], t[N];
int n, m, ans;
void solve() {
  memcpy(he.c, t, sizeof(char) * N);
  he.ini(m, 0);
  memset(f, 0, sizeof(f));
  for (int i = 1, now = 0, las = 1; i <= n; i++, swap(now, las)) {
    memset(f[now], 0, sizeof(f[now]));
    if (c[0][i] == t[1]) addx(f[now][1][0], 1);
    if (c[1][i] == t[1]) addx(f[now][1][1], 1);
    for (int j = 0; j <= m; j++) {
      for (int k = 0; k <= 1; k++) {
        if (!f[las][j][k]) continue;
        if (c[k][i] == t[j + 1]) {
          addx(f[now][j + 1][k], f[las][j][k]);
          if (c[k ^ 1][i] == t[j + 2]) addx(f[now][j + 2][k ^ 1], f[las][j][k]);
        }
      }
    }
    addx(ans, add(f[now][m][0], f[now][m][1]));
    for (int j = 1; j <= m / 2; j++) {
      int cando1 = 0, cando2 = 0, cando3 = 0, cando4 = 0;
      if (i >= j && ha[0][1].qry(i, i - j + 1) == he.qry(1, j) &&
          ha[1][0].qry(i - j + 1, i) == he.qry(j + 1, 2 * j))
        addx(f[now][j * 2][1], 1), cando1 = 1;
      if (i >= j && ha[1][1].qry(i, i - j + 1) == he.qry(1, j) &&
          ha[0][0].qry(i - j + 1, i) == he.qry(j + 1, 2 * j))
        addx(f[now][j * 2][0], 1), cando2 = 1;
      if (j > 1 && i <= n - j + 1 &&
          ha[0][0].qry(i, i + j - 1) == he.qry(m - 2 * j + 1, m - j) &&
          ha[1][1].qry(i + j - 1, i) == he.qry(m - j + 1, m))
        addx(ans, f[las][m - 2 * j][0]), cando3 = 1;
      if (j > 1 && i <= n - j + 1 &&
          ha[1][0].qry(i, i + j - 1) == he.qry(m - 2 * j + 1, m - j) &&
          ha[0][1].qry(i + j - 1, i) == he.qry(m - j + 1, m))
        addx(ans, f[las][m - 2 * j][1]), cando4 = 1;
      if (j * 2 == m) addx(ans, mul(cando1 + cando2 + cando3 + cando4, inv2));
    }
  }
}
int vis[2][N];
void dfs1(int x, int y, int len) {
  if (c[x][y] != t[len] || vis[x][y]) return;
  if (len == m) {
    ++ans;
    return;
  }
  vis[x][y] = 1;
  if (!x)
    dfs1(x + 1, y, len + 1);
  else
    dfs1(x - 1, y, len + 1);
  if (y != 1) dfs1(x, y - 1, len + 1);
  if (y != n) dfs1(x, y + 1, len + 1);
  vis[x][y] = 0;
}
int main() {
  scanf("%s%s%s", c[0] + 1, c[1] + 1, t + 1);
  n = strlen(c[0] + 1);
  m = strlen(t + 1);
  if (m <= 6) {
    for (int i = 0; i <= 1; i++) {
      for (int j = 1; j <= n; j++) {
        dfs1(i, j, 1);
      }
    }
    print(ans, '\n');
    return 0;
  }
  memcpy(ha[0][0].c, c[0], sizeof(char) * N);
  memcpy(ha[0][1].c, c[0], sizeof(char) * N);
  reverse(ha[0][1].c + 1, ha[0][1].c + n + 1);
  memcpy(ha[1][0].c, c[1], sizeof(char) * N);
  memcpy(ha[1][1].c, c[1], sizeof(char) * N);
  reverse(ha[1][1].c + 1, ha[1][1].c + n + 1);
  ha[0][0].ini(n, 0);
  ha[0][1].ini(n, 1);
  ha[1][0].ini(n, 0);
  ha[1][1].ini(n, 1);
  solve();
  reverse(t + 1, t + m + 1);
  solve();
  print(ans, '\n');
  return 0;
}
