#include <bits/stdc++.h>
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;
const int INF = 0x7f7f7f7f;
const double pi = acos(-1.0);
const double EPS = 1e-6;
using namespace std;
char s[MAXN];
int k;
int main() {
  scanf("%s%d", s, &k);
  int len = strlen(s);
  if (len < k || len % k != 0) {
    printf("NO\n");
    return 0;
  }
  len /= k;
  bool ans = true;
  for (int i = 0; i < k; ++i) {
    bool flag = true;
    for (int j = 0; j < len / 2; ++j)
      if (s[j + i * len] != s[len - j - 1 + i * len]) {
        flag = false;
        break;
      }
    if (!flag) {
      ans = false;
      break;
    }
  }
  if (ans)
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}

### Prompt

Your challenge is to write a cpp solution 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>
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;
const int INF = 0x7f7f7f7f;
const double pi = acos(-1.0);
const double EPS = 1e-6;
using namespace std;
char s[MAXN];
int k;
int main() {
  scanf("%s%d", s, &k);
  int len = strlen(s);
  if (len < k || len % k != 0) {
    printf("NO\n");
    return 0;
  }
  len /= k;
  bool ans = true;
  for (int i = 0; i < k; ++i) {
    bool flag = true;
    for (int j = 0; j < len / 2; ++j)
      if (s[j + i * len] != s[len - j - 1 + i * len]) {
        flag = false;
        break;
      }
    if (!flag) {
      ans = false;
      break;
    }
  }
  if (ans)
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}

```