Problem: 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".

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mod = 10007;
inline int add(int a, int b) {
  if ((a += b) >= mod) a -= mod;
  return a;
}
inline int mult(int a, int b) { return a * b % mod; }
inline int dec(int a, int b) {
  if ((a -= b) < 0) a += mod;
  return a;
}
inline int power(int a, int b) {
  int out = 1;
  while (b) {
    if (b & 1) out = mult(out, a);
    a = mult(a, a);
    b >>= 1;
  }
  return out;
}
namespace polynomial {
vector<int> Mult(vector<int> A, vector<int> B, int n, int m) {
  vector<int> ret(n + m - 1, 0);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) ret[i + j] = add(ret[i + j], mult(A[i], B[j]));
  return ret;
}
vector<int> Mult(vector<int> A, vector<int> B) {
  return Mult(A, B, A.size(), B.size());
}
vector<int> Add(vector<int> A, vector<int> B) {
  vector<int> ret;
  ret.resize(max(A.size(), B.size()));
  for (int i = 0; i < ret.size(); i++) {
    if (i < A.size()) ret[i] = A[i];
    if (i < B.size()) ret[i] = add(ret[i], B[i]);
  }
  return ret;
}
vector<int> inverse(vector<int> f, int len) {
  vector<int> ret;
  if (len == 1) {
    ret.push_back(power(f[0], mod - 2));
    return ret;
  }
  vector<int> h = inverse(f, (len + 1) >> 1);
  int tlen = 1;
  while (tlen < (len << 1)) tlen <<= 1;
  ret.resize(tlen);
  vector<int> h2 = Mult(Mult(h, h), f);
  h2.resize(tlen);
  h.resize(tlen);
  for (int i = 0; i < tlen; i++) ret[i] = dec(mult(2, h[i]), h2[i]);
  ret.resize(len);
  return ret;
}
vector<int> inverse(vector<int> A) { return inverse(A, A.size()); }
vector<int> derivative(vector<int> A) {
  vector<int> ret;
  ret.resize(A.size() - 1);
  for (int i = 1; i < A.size(); i++) ret[i - 1] = mult(i, A[i]);
  return ret;
}
vector<int> integral(vector<int> A) {
  vector<int> ret;
  ret.resize(A.size() + 1);
  for (int i = 0; i < A.size(); i++)
    ret[i + 1] = mult(A[i], power(i + 1, mod - 2));
  ret[0] = 0;
  return ret;
}
vector<int> ln(vector<int> A, int len) {
  A = integral(Mult(inverse(A), derivative(A)));
  A.resize(len);
  return A;
}
vector<int> ln(vector<int> f) { return ln(f, f.size()); }
vector<int> exp(vector<int> f, int n) {
  vector<int> ret;
  if (n == 1) {
    ret.push_back(1);
    return ret;
  }
  vector<int> h = exp(f, (n + 1) >> 1);
  h.resize(n);
  vector<int> l = ln(h);
  ret.resize(n);
  l[0] = dec(add(f[0], 1), l[0]);
  for (int i = 1; i < n; i++) l[i] = dec(f[i], l[i]);
  ret = Mult(h, l);
  ret.resize(n);
  return ret;
}
vector<int> exp(vector<int> f) { return exp(f, f.size()); }
vector<int> Power(vector<int> f, int k) {
  int a = -1, b = -1;
  for (int i = 0; i < f.size(); i++) {
    if (f[i] != 0) {
      a = f[i];
      b = i;
      break;
    }
  }
  if (a == -1) return f;
  int ia = power(a, mod - 2);
  vector<int> g(f.size());
  if (b * k >= g.size()) {
    for (int i = 0; i < g.size(); i++) g[i] = 0;
    return g;
  }
  for (int i = b; i < f.size(); i++) g[i - b] = mult(f[i], ia);
  g = ln(g);
  for (int i = 0; i < g.size(); i++) g[i] = mult(g[i], k);
  g = exp(g);
  a = power(a, k);
  for (int i = 0; i < b * k; i++) f[i] = 0;
  for (int i = b * k; i < g.size(); i++) f[i] = mult(g[i - b * k], a);
  return f;
}
pair<vector<int>, vector<int> > division(vector<int> f, vector<int> g, int n,
                                         int m) {
  f.resize(n + 1);
  g.resize(m + 1);
  vector<int> rf = f, rg = g, q;
  reverse(rf.begin(), rf.end());
  reverse(rg.begin(), rg.end());
  rf.resize(n - m + 1);
  rg = inverse(rg, n - m + 1);
  q = Mult(rf, rg);
  q.resize(n - m + 1);
  reverse(q.begin(), q.end());
  vector<int> r, tmp = Mult(q, g);
  r.resize(m);
  for (int i = 0; i < m; i++) r[i] = dec(f[i], tmp[i]);
  return make_pair(q, r);
}
inline pair<vector<int>, vector<int> > division(vector<int> f, vector<int> g) {
  if (f.size() < g.size()) return make_pair(vector<int>(1, 0), f);
  return division(f, g, f.size() - 1, g.size() - 1);
}
inline vector<int> getremainder(vector<int> f, vector<int> g) {
  if (f.size() < g.size()) return f;
  return division(f, g, f.size() - 1, g.size() - 1).second;
}
}  // namespace polynomial
using namespace polynomial;
namespace linearrecursion {
vector<int> c, a;
inline int linearrecursion(vector<int> g, vector<int> st, int n) {
  int k = g.size();
  reverse(g.begin(), g.end());
  g.push_back(1);
  for (int i = 0; i < k; i++) g[i] = dec(0, g[i]);
  c.push_back(1);
  a.push_back(0);
  a.push_back(1);
  while (n) {
    if (n & 1) c = getremainder(Mult(c, a), g);
    a = getremainder(Mult(a, a), g);
    n >>= 1;
  }
  int ans = 0;
  c.resize(k);
  for (int i = 0; i < k; i++) ans = add(ans, mult(st[i], c[i]));
  return ans;
}
}  // namespace linearrecursion
namespace BM {
vector<int> seq[3010];
int fail[3010], cnt = 0, delta[3010];
inline vector<int> BM(vector<int> P, int n) {
  vector<int> R1;
  cnt = 0;
  seq[0].clear();
  for (int i = 0; i < n; i++) {
    int tval = 0;
    for (int j = 0; j < seq[cnt].size(); j++)
      tval = add(tval, mult(seq[cnt][j], P[i - j - 1]));
    delta[i] = dec(P[i], tval);
    if (tval == P[i]) continue;
    fail[cnt] = i;
    ++cnt;
    seq[cnt].clear();
    if (cnt == 1)
      for (int j = 0; j <= i; j++) seq[cnt].push_back(0);
    else {
      R1.clear();
      int mxpos = cnt - 2;
      for (int j = cnt - 3; j >= 0; j--)
        if ((int)seq[mxpos].size() - fail[mxpos] > (int)seq[j].size() - fail[j])
          mxpos = j;
      int tmp = mult(delta[i], power(delta[fail[mxpos]], mod - 2));
      for (int j = 0; j < i - fail[mxpos] - 1; j++) R1.push_back(0);
      R1.push_back(tmp);
      for (int j = 0; j < seq[mxpos].size(); j++)
        R1.push_back(dec(0, mult(tmp, seq[mxpos][j])));
      seq[cnt] = Add(seq[cnt - 1], R1);
    }
  }
  return seq[cnt];
}
}  // namespace BM
char s[210];
int n, f[610][210][210], done[610];
const int maxlen = 610;
int main() {
  scanf("%s%d", s + 1, &n);
  int m = strlen(s + 1);
  f[0][1][m] = 1;
  for (int i = 1; i < maxlen; i++) {
    done[i] = mult(done[i - 1], 26);
    for (int j = 0; j <= m; j++) {
      for (int k = m + 1; k >= j; k--) {
        if (s[j] == s[k]) {
          if (j + 1 <= k - 1)
            f[i][j + 1][k - 1] = add(f[i][j + 1][k - 1], f[i - 1][j][k]);
          else
            done[i] = add(done[i], f[i - 1][j][k]);
          f[i][j][k] = add(f[i][j][k], mult(f[i - 1][j][k], 25));
        } else {
          if (j == k)
            done[i] = add(done[i], mult(f[i - 1][j][k], 2));
          else {
            f[i][j + 1][k] = add(f[i][j + 1][k], f[i - 1][j][k]);
            f[i][j][k - 1] = add(f[i][j][k - 1], f[i - 1][j][k]);
          }
          f[i][j][k] = add(f[i][j][k], mult(f[i - 1][j][k], 24));
        }
      }
    }
  }
  vector<int> seqs(maxlen << 1);
  for (int i = m; i < (maxlen << 1); i++) {
    if (i & 1) {
      seqs[i] = mult(done[i >> 1], 26);
      for (int j = 1; j <= m; j++) seqs[i] = add(seqs[i], f[i >> 1][j][j]);
    } else {
      seqs[i] = done[i >> 1];
    }
  }
  printf("%d\n", linearrecursion::linearrecursion(BM::BM(seqs, maxlen << 1),
                                                  seqs, n + m));
  return 0;
}