#include <bits/stdc++.h>
#pragma comment(linker, "/stack:20000000")
using namespace std;
double __begin;
template <typename T1, typename T2, typename T3>
struct triple {
  T1 a;
  T2 b;
  T3 c;
  triple(){};
  triple(T1 _a, T2 _b, T3 _c) : a(_a), b(_b), c(_c) {}
};
template <typename T1, typename T2, typename T3>
bool operator<(const triple<T1, T2, T3> &t1, const triple<T1, T2, T3> &t2) {
  if (t1.a != t2.a)
    return t1.a < t2.a;
  else
    return t1.b < t2.b;
}
template <typename T1, typename T2, typename T3>
inline std::ostream &operator<<(std::ostream &os, const triple<T1, T2, T3> &t) {
  return os << "(" << t.a << ", " << t.b << ", " << t.c << ")";
}
inline int bits_count(int v) {
  v = v - ((v >> 1) & 0x55555555);
  v = (v & 0x33333333) + ((v >> 2) & 0x33333333);
  return ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
}
inline int bits_count(long long v) {
  int t = v >> 32;
  int p = (v & ((1LL << 32) - 1));
  return bits_count(t) + bits_count(p);
}
unsigned int reverse_bits(register unsigned int x) {
  x = (((x & 0xaaaaaaaa) >> 1) | ((x & 0x55555555) << 1));
  x = (((x & 0xcccccccc) >> 2) | ((x & 0x33333333) << 2));
  x = (((x & 0xf0f0f0f0) >> 4) | ((x & 0x0f0f0f0f) << 4));
  x = (((x & 0xff00ff00) >> 8) | ((x & 0x00ff00ff) << 8));
  return ((x >> 16) | (x << 16));
}
inline int sign(int x) { return (x >> 31) | (-x >> 31); }
inline bool ispow2(int x) { return (x != 0 && (x & (x - 1)) == 0); }
template <typename T1, typename T2>
inline std::ostream &operator<<(std::ostream &os, const std::pair<T1, T2> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
inline std::ostream &operator<<(std::ostream &os, const std::vector<T> &v) {
  bool first = true;
  os << "[";
  for (unsigned int i = 0; i < v.size(); i++) {
    if (!first) os << ", ";
    os << v[i];
    first = false;
  }
  return os << "]";
}
template <typename T>
inline std::ostream &operator<<(std::ostream &os, const std::set<T> &v) {
  bool first = true;
  os << "[";
  for (typename std::set<T>::const_iterator ii = v.begin(); ii != v.end();
       ++ii) {
    if (!first) os << ", ";
    os << *ii;
    first = false;
  }
  return os << "]";
}
template <typename T1, typename T2>
inline std::ostream &operator<<(std::ostream &os, const std::map<T1, T2> &v) {
  bool first = true;
  os << "[";
  for (typename std::map<T1, T2>::const_iterator ii = v.begin(); ii != v.end();
       ++ii) {
    if (!first) os << ", ";
    os << *ii;
    first = false;
  }
  return os << "]";
}
template <typename T, typename T2>
void printarray(T a[], T2 sz, T2 beg = 0) {
  for (T2 i = beg; i < sz; i++) cout << a[i] << " ";
  cout << endl;
}
inline long long binpow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res *= x;
    x *= x;
    n >>= 1;
  }
  return res;
}
inline long long powmod(long long x, long long n, long long _mod) {
  long long res = 1;
  while (n) {
    if (n & 1) res = (res * x) % _mod;
    x = (x * x) % _mod;
    n >>= 1;
  }
  return res;
}
inline long long gcd(long long a, long long b) {
  long long t;
  while (b) {
    a = a % b;
    t = a;
    a = b;
    b = t;
  }
  return a;
}
inline int gcd(int a, int b) {
  int t;
  while (b) {
    a = a % b;
    t = a;
    a = b;
    b = t;
  }
  return a;
}
inline long long lcm(int a, int b) { return a / gcd(a, b) * (long long)b; }
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long gcd(long long a, long long b, long long c) {
  return gcd(gcd(a, b), c);
}
inline int gcd(int a, int b, int c) { return gcd(gcd(a, b), c); }
inline long long lcm(long long a, long long b, long long c) {
  return lcm(lcm(a, b), c);
}
inline long long lcm(int a, int b, int c) {
  return lcm(lcm(a, b), (long long)c);
}
inline long long max(long long a, long long b) { return (a > b) ? a : b; }
inline int max(int a, int b) { return (a > b) ? a : b; }
inline double max(double a, double b) { return (a > b) ? a : b; }
inline long long max(long long a, long long b, long long c) {
  return max(a, max(b, c));
}
inline int max(int a, int b, int c) { return max(a, max(b, c)); }
inline double max(double a, double b, double c) { return max(a, max(b, c)); }
inline long long min(long long a, long long b) { return (a < b) ? a : b; }
inline int min(int a, int b) { return (a < b) ? a : b; }
inline double min(double a, double b) { return (a < b) ? a : b; }
inline long long min(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
inline int min(int a, int b, int c) { return min(a, min(b, c)); }
inline double min(double a, double b, double c) { return min(a, min(b, c)); }
template <class T>
inline void getar(T a, int n, int m) {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; ++j) {
      scanf("%d", &a[i][j]);
    }
}
inline void getar(int *a, int n) {
  for (int ii = 0; ii < n; ii++) {
    scanf("%d", a + ii);
  }
}
inline void getar(pair<int, int> *a, int n) {
  for (int ii = 0; ii < n; ii++) {
    scanf("%d%d", &a[ii].first, &a[ii].second);
  }
}
inline void getar(long long *a, int n) {
  for (int ii = 0; ii < n; ii++) {
    scanf("%I64d", a + ii);
  }
}
template <class T>
inline void cinarr(T a, int n) {
  for (int i = 0; i < n; ++i) cin >> a[i];
}
template <class T>
inline bool scan_d(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
inline void out(int x) {
  if (x > 9) out(x / 10);
  putchar(x % 10 + '0');
}
int n, m, dp[2][501][501];
char land[501][501];
bool isRight(int x, int y) {
  if (x >= 0 && x < n && y >= 0 && y < m) return true;
  return false;
}
void update(int &x, int y) {
  x += y;
  x %= 1000000007;
}
void Add(int x1, int y1, int x2, int y2, int px1, int px2, int flag) {
  if (isRight(x1, y1) && isRight(x2, y2) && land[x1][y1] == land[x2][y2])
    update(dp[flag ^ 1][x1][x2], dp[flag][px1][px2]);
}
int main() {
  while (scanf("%d%d", &n, &m) != EOF) {
    for (int i = 0; i < n; i++) {
      scanf("%s", land[i]);
    }
    int flag = 0, sn = (n + m - 2) / 2;
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < n; i++)
      for (int j = i; j < n; j++) {
        int x1 = i, x2 = j, y1 = sn - x1, y2 = m + n - 2 - sn - x2;
        if (isRight(x1, y1) && isRight(x2, y2) &&
            abs(x1 - x2) + abs(y1 - y2) <= 1) {
          dp[flag][i][j] = (land[x1][y1] == land[x2][y2]) ? 1 : 0;
        }
      }
    for (int step = sn; step >= 1; step--, flag = flag ^ 1) {
      memset(dp[flag ^ 1], 0, sizeof(dp[flag ^ 1]));
      for (int i = 0; i < n; i++)
        for (int j = i; j < n; j++) {
          int x1 = i, x2 = j, y1 = step - x1, y2 = m + n - 2 - step - x2;
          Add(x1, y1 - 1, x2, y2 + 1, x1, x2, flag);
          Add(x1, y1 - 1, x2 + 1, y2, x1, x2, flag);
          Add(x1 - 1, y1, x2, y2 + 1, x1, x2, flag);
          Add(x1 - 1, y1, x2 + 1, y2, x1, x2, flag);
        }
    }
    printf("%d\n", dp[flag][0][n - 1]);
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

Peppa the Pig was walking and walked into the forest. What a strange coincidence! The forest has the shape of a rectangle, consisting of n rows and m columns. We enumerate the rows of the rectangle from top to bottom with numbers from 1 to n, and the columns — from left to right with numbers from 1 to m. Let's denote the cell at the intersection of the r-th row and the c-th column as (r, c).

Initially the pig stands in cell (1, 1), and in the end she wants to be in cell (n, m). Since the pig is in a hurry to get home, she can go from cell (r, c), only to either cell (r + 1, c) or (r, c + 1). She cannot leave the forest.

The forest, where the pig is, is very unusual. Some cells of the forest similar to each other, and some look very different. Peppa enjoys taking pictures and at every step she takes a picture of the cell where she is now. The path through the forest is considered to be beautiful if photographs taken on her way, can be viewed in both forward and in reverse order, showing the same sequence of photos. More formally, the line formed by the cells in order of visiting should be a palindrome (you can read a formal definition of a palindrome in the previous problem).

Count the number of beautiful paths from cell (1, 1) to cell (n, m). Since this number can be very large, determine the remainder after dividing it by 109 + 7.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 500) — the height and width of the field.

Each of the following n lines contains m lowercase English letters identifying the types of cells of the forest. Identical cells are represented by identical letters, different cells are represented by different letters.

Output

Print a single integer — the number of beautiful paths modulo 109 + 7.

Examples

Input

3 4
aaab
baaa
abba


Output

3

Note

Picture illustrating possibilities for the sample test. 

<image>

<image>

<image>

### Response

```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/stack:20000000")
using namespace std;
double __begin;
template <typename T1, typename T2, typename T3>
struct triple {
  T1 a;
  T2 b;
  T3 c;
  triple(){};
  triple(T1 _a, T2 _b, T3 _c) : a(_a), b(_b), c(_c) {}
};
template <typename T1, typename T2, typename T3>
bool operator<(const triple<T1, T2, T3> &t1, const triple<T1, T2, T3> &t2) {
  if (t1.a != t2.a)
    return t1.a < t2.a;
  else
    return t1.b < t2.b;
}
template <typename T1, typename T2, typename T3>
inline std::ostream &operator<<(std::ostream &os, const triple<T1, T2, T3> &t) {
  return os << "(" << t.a << ", " << t.b << ", " << t.c << ")";
}
inline int bits_count(int v) {
  v = v - ((v >> 1) & 0x55555555);
  v = (v & 0x33333333) + ((v >> 2) & 0x33333333);
  return ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
}
inline int bits_count(long long v) {
  int t = v >> 32;
  int p = (v & ((1LL << 32) - 1));
  return bits_count(t) + bits_count(p);
}
unsigned int reverse_bits(register unsigned int x) {
  x = (((x & 0xaaaaaaaa) >> 1) | ((x & 0x55555555) << 1));
  x = (((x & 0xcccccccc) >> 2) | ((x & 0x33333333) << 2));
  x = (((x & 0xf0f0f0f0) >> 4) | ((x & 0x0f0f0f0f) << 4));
  x = (((x & 0xff00ff00) >> 8) | ((x & 0x00ff00ff) << 8));
  return ((x >> 16) | (x << 16));
}
inline int sign(int x) { return (x >> 31) | (-x >> 31); }
inline bool ispow2(int x) { return (x != 0 && (x & (x - 1)) == 0); }
template <typename T1, typename T2>
inline std::ostream &operator<<(std::ostream &os, const std::pair<T1, T2> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
inline std::ostream &operator<<(std::ostream &os, const std::vector<T> &v) {
  bool first = true;
  os << "[";
  for (unsigned int i = 0; i < v.size(); i++) {
    if (!first) os << ", ";
    os << v[i];
    first = false;
  }
  return os << "]";
}
template <typename T>
inline std::ostream &operator<<(std::ostream &os, const std::set<T> &v) {
  bool first = true;
  os << "[";
  for (typename std::set<T>::const_iterator ii = v.begin(); ii != v.end();
       ++ii) {
    if (!first) os << ", ";
    os << *ii;
    first = false;
  }
  return os << "]";
}
template <typename T1, typename T2>
inline std::ostream &operator<<(std::ostream &os, const std::map<T1, T2> &v) {
  bool first = true;
  os << "[";
  for (typename std::map<T1, T2>::const_iterator ii = v.begin(); ii != v.end();
       ++ii) {
    if (!first) os << ", ";
    os << *ii;
    first = false;
  }
  return os << "]";
}
template <typename T, typename T2>
void printarray(T a[], T2 sz, T2 beg = 0) {
  for (T2 i = beg; i < sz; i++) cout << a[i] << " ";
  cout << endl;
}
inline long long binpow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res *= x;
    x *= x;
    n >>= 1;
  }
  return res;
}
inline long long powmod(long long x, long long n, long long _mod) {
  long long res = 1;
  while (n) {
    if (n & 1) res = (res * x) % _mod;
    x = (x * x) % _mod;
    n >>= 1;
  }
  return res;
}
inline long long gcd(long long a, long long b) {
  long long t;
  while (b) {
    a = a % b;
    t = a;
    a = b;
    b = t;
  }
  return a;
}
inline int gcd(int a, int b) {
  int t;
  while (b) {
    a = a % b;
    t = a;
    a = b;
    b = t;
  }
  return a;
}
inline long long lcm(int a, int b) { return a / gcd(a, b) * (long long)b; }
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long gcd(long long a, long long b, long long c) {
  return gcd(gcd(a, b), c);
}
inline int gcd(int a, int b, int c) { return gcd(gcd(a, b), c); }
inline long long lcm(long long a, long long b, long long c) {
  return lcm(lcm(a, b), c);
}
inline long long lcm(int a, int b, int c) {
  return lcm(lcm(a, b), (long long)c);
}
inline long long max(long long a, long long b) { return (a > b) ? a : b; }
inline int max(int a, int b) { return (a > b) ? a : b; }
inline double max(double a, double b) { return (a > b) ? a : b; }
inline long long max(long long a, long long b, long long c) {
  return max(a, max(b, c));
}
inline int max(int a, int b, int c) { return max(a, max(b, c)); }
inline double max(double a, double b, double c) { return max(a, max(b, c)); }
inline long long min(long long a, long long b) { return (a < b) ? a : b; }
inline int min(int a, int b) { return (a < b) ? a : b; }
inline double min(double a, double b) { return (a < b) ? a : b; }
inline long long min(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
inline int min(int a, int b, int c) { return min(a, min(b, c)); }
inline double min(double a, double b, double c) { return min(a, min(b, c)); }
template <class T>
inline void getar(T a, int n, int m) {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; ++j) {
      scanf("%d", &a[i][j]);
    }
}
inline void getar(int *a, int n) {
  for (int ii = 0; ii < n; ii++) {
    scanf("%d", a + ii);
  }
}
inline void getar(pair<int, int> *a, int n) {
  for (int ii = 0; ii < n; ii++) {
    scanf("%d%d", &a[ii].first, &a[ii].second);
  }
}
inline void getar(long long *a, int n) {
  for (int ii = 0; ii < n; ii++) {
    scanf("%I64d", a + ii);
  }
}
template <class T>
inline void cinarr(T a, int n) {
  for (int i = 0; i < n; ++i) cin >> a[i];
}
template <class T>
inline bool scan_d(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
inline void out(int x) {
  if (x > 9) out(x / 10);
  putchar(x % 10 + '0');
}
int n, m, dp[2][501][501];
char land[501][501];
bool isRight(int x, int y) {
  if (x >= 0 && x < n && y >= 0 && y < m) return true;
  return false;
}
void update(int &x, int y) {
  x += y;
  x %= 1000000007;
}
void Add(int x1, int y1, int x2, int y2, int px1, int px2, int flag) {
  if (isRight(x1, y1) && isRight(x2, y2) && land[x1][y1] == land[x2][y2])
    update(dp[flag ^ 1][x1][x2], dp[flag][px1][px2]);
}
int main() {
  while (scanf("%d%d", &n, &m) != EOF) {
    for (int i = 0; i < n; i++) {
      scanf("%s", land[i]);
    }
    int flag = 0, sn = (n + m - 2) / 2;
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < n; i++)
      for (int j = i; j < n; j++) {
        int x1 = i, x2 = j, y1 = sn - x1, y2 = m + n - 2 - sn - x2;
        if (isRight(x1, y1) && isRight(x2, y2) &&
            abs(x1 - x2) + abs(y1 - y2) <= 1) {
          dp[flag][i][j] = (land[x1][y1] == land[x2][y2]) ? 1 : 0;
        }
      }
    for (int step = sn; step >= 1; step--, flag = flag ^ 1) {
      memset(dp[flag ^ 1], 0, sizeof(dp[flag ^ 1]));
      for (int i = 0; i < n; i++)
        for (int j = i; j < n; j++) {
          int x1 = i, x2 = j, y1 = step - x1, y2 = m + n - 2 - step - x2;
          Add(x1, y1 - 1, x2, y2 + 1, x1, x2, flag);
          Add(x1, y1 - 1, x2 + 1, y2, x1, x2, flag);
          Add(x1 - 1, y1, x2, y2 + 1, x1, x2, flag);
          Add(x1 - 1, y1, x2 + 1, y2, x1, x2, flag);
        }
    }
    printf("%d\n", dp[flag][0][n - 1]);
  }
  return 0;
}

```