Mr. Kitayuta has kindly given you a string s consisting of lowercase English letters. You are asked to insert exactly n lowercase English letters into s to make it a palindrome. (A palindrome is a string that reads the same forward and backward. For example, "noon", "testset" and "a" are all palindromes, while "test" and "kitayuta" are not.) You can choose any n lowercase English letters, and insert each of them to any position of s, possibly to the beginning or the end of s. You have to insert exactly n letters even if it is possible to turn s into a palindrome by inserting less than n letters.

Find the number of the palindromes that can be obtained in this way, modulo 10007.

Input

The first line contains a string s (1 ≤ |s| ≤ 200). Each character in s is a lowercase English letter.

The second line contains an integer n (1 ≤ n ≤ 109).

Output

Print the number of the palindromes that can be obtained, modulo 10007.

Examples

Input

revive
1


Output

1


Input

add
2


Output

28

Note

For the first sample, you can obtain the palindrome "reviver" by inserting 'r' to the end of "revive".

For the second sample, the following 28 palindromes can be obtained: "adada", "adbda", ..., "adzda", "dadad" and "ddadd".
#include <bits/stdc++.h>
using namespace std;
const int M = 10007;
const int MaxNL = 200;
int n, nL, m, nR, nG, nB, len;
char s[MaxNL + 5];
int f[MaxNL + 1][MaxNL + 1][MaxNL + 1];
__attribute__((optimize("-O2"))) int solve(const int &nR, const int &l,
                                           const int &r) {
  if (nR < 0) return 0;
  int &res = f[nR][l][r];
  if (res != -1) return res;
  if (l == r || (r == l + 1 && s[l] == s[r]))
    res = nR == 0;
  else if (s[l] == s[r])
    res = solve(nR, l + 1, r - 1);
  else {
    int wL = solve(nR - 1, l + 1, r);
    int wR = solve(nR - 1, l, r - 1);
    if ((res = wL + wR) >= M) res -= M;
  }
  return res;
}
struct matrix {
  int a[MaxNL * 2][MaxNL * 2];
  __attribute__((optimize("-O2"))) __inline__
      __attribute__((always_inline)) void
      init() {
    for (int i = 1; i <= m; ++i) {
      fill(a[i] + 1, a[i] + m + 1, 0);
      a[i][i] = 1;
    }
  }
  __attribute__((optimize("-O2"))) friend __inline__
      __attribute__((always_inline)) matrix
      operator*(const matrix &lhs, const matrix &rhs) {
    matrix res;
    for (int i = 1; i <= m; ++i)
      for (int j = 1; j <= m; ++j) {
        long long w = 0;
        for (int k = i; k <= j; ++k) w += lhs.a[i][k] * rhs.a[k][j];
        res.a[i][j] = w % M;
      }
    return res;
  }
  __attribute__((optimize("-O2"))) __inline__
      __attribute__((always_inline)) void
      operator=(const matrix &rhs) {
    for (int i = 1; i <= m; ++i) copy(rhs.a[i] + 1, rhs.a[i] + m + 1, a[i] + 1);
  }
};
matrix adj, u, v;
__attribute__((optimize("-O2"))) __inline__ __attribute__((always_inline))
matrix
matpow(matrix a, const int &n) {
  int k, t = 233333;
  for (int l = 2; l <= 64; ++l) {
    int time = 0;
    for (int i = n; i; i /= l) time += l - 1 + (i % l != 0);
    if (time < t) t = time, k = l;
  }
  matrix res;
  res.init();
  for (int i = n; i; i /= k) {
    matrix powT = a;
    for (int j = 2; j <= k; ++j) {
      powT = powT * a;
      if (j == i % k) res = res * powT;
    }
    if (i % k == 1) res = res * a;
    a = powT;
  }
  return res;
}
__attribute__((optimize("-O2"))) int main() {
  scanf("%s%d", s, &n);
  n += (nL = strlen(s));
  nR = nL - 1, nG = nB = nL + 1 >> 1;
  m = nR + nG + nB;
  len = n + 1 >> 1;
  for (int i = 0; i <= nR; ++i)
    for (int j = 0; j < nL; ++j) fill(f[i][j] + j, f[i][j] + nL, -1);
  for (int i = 1; i <= nR; ++i) adj.a[i][i] = 24;
  for (int i = nR + 1; i <= nR + nG; ++i) adj.a[i][i] = 25;
  for (int i = nR + nG + 1; i <= m; ++i) adj.a[i][i] = 26;
  for (int i = 1; i < nR + nG; ++i) adj.a[i][i + 1] = 1;
  for (int i = nR + 1; i <= nR + nG; ++i) adj.a[i][i + nG] = 1;
  u = matpow(adj, len - 1);
  v = u * adj;
  long long outcome = 0;
  for (int nR = 0; nR <= ::nR; ++nR) {
    int nG = (nL - nR + 1) >> 1;
    int nB = len - nR - nG;
    if (nB < 0) continue;
    int w = solve(nR, 0, nL - 1);
    outcome += w * v.a[::nR - nR + 1][::nR + ::nG + nG];
    if ((n & 1) && (nL - nR & 1 ^ 1))
      outcome -= w * u.a[::nR - nR + 1][::nR + nG];
  }
  cout << (outcome % M + M) % M;
  return 0;
}
