#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7;
inline int add(int a, int b) {
  if ((a += b) >= P) a -= P;
  return a;
}
inline int sub(int a, int b) {
  if ((a -= b) < 0) a += P;
  return a;
}
inline int mul(int a, int b) { return 1ll * a * b % P; }
inline int kpow(int a, int b) {
  int r = 1;
  for (; b; b >>= 1, a = mul(a, a)) {
    if (b & 1) r = mul(r, a);
  }
  return r;
}
const int N = 1010;
int n, m;
char s[N];
int main() {
  scanf("%s%d", s, &m);
  n = strlen(s);
  if (n % m) {
    return 0 * puts("NO");
  }
  bool ans = 1;
  int d = n / m;
  for (int l = 0, r = d; l < n; l = r, r += d) {
    int tl = l, tr = r - 1;
    bool ok = 1;
    for (int i = (0); i < (d); i++) {
      ok &= s[tl] == s[tr];
      tl++, tr--;
    }
    ans &= ok;
  }
  puts(ans ? "YES" : "NO");
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string s.

<image>

He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly k messages in his own bag, each was a palindrome string and all those strings had the same length.

He asked you to help him and tell him if he has worn his own back-bag. Check if the given string s is a concatenation of k palindromes of the same length.

Input

The first line of input contains string s containing lowercase English letters (1 ≤ |s| ≤ 1000).

The second line contains integer k (1 ≤ k ≤ 1000).

Output

Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.

Examples

Input

saba
2


Output

NO


Input

saddastavvat
2


Output

YES

Note

Palindrome is a string reading the same forward and backward.

In the second sample, the faxes in his back-bag can be "saddas" and "tavvat".

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7;
inline int add(int a, int b) {
  if ((a += b) >= P) a -= P;
  return a;
}
inline int sub(int a, int b) {
  if ((a -= b) < 0) a += P;
  return a;
}
inline int mul(int a, int b) { return 1ll * a * b % P; }
inline int kpow(int a, int b) {
  int r = 1;
  for (; b; b >>= 1, a = mul(a, a)) {
    if (b & 1) r = mul(r, a);
  }
  return r;
}
const int N = 1010;
int n, m;
char s[N];
int main() {
  scanf("%s%d", s, &m);
  n = strlen(s);
  if (n % m) {
    return 0 * puts("NO");
  }
  bool ans = 1;
  int d = n / m;
  for (int l = 0, r = d; l < n; l = r, r += d) {
    int tl = l, tr = r - 1;
    bool ok = 1;
    for (int i = (0); i < (d); i++) {
      ok &= s[tl] == s[tr];
      tl++, tr--;
    }
    ans &= ok;
  }
  puts(ans ? "YES" : "NO");
  return 0;
}

```