Mike and Fax

A. Mike and Fax

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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.



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.

Sample test(s)

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".

题意：

     是否存在从头到尾k份相同的字符回文串。不在状态所以一直没弄懂什么意思，一直Wa。。。

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdio>
#include<vector>
using namespace std;
typedef __int64 ll;
#define CRL(a) memset(a,0,sizeof(a))
#define T 10100
char s[1005];
int c[1005];
bool find(int i,int j)
{
while(i<j)
{
if(s[i]==s[j])i++,j--;
else break;
}
if(i>=j)return true;
return false;
}
int main()
{
/*freopen("input.txt","r",stdin);*/
int i,k,flag;
while(~scanf("%s",&s))
{
CRL(c);
scanf("%d",&k);flag=1;
int len = strlen(s);
if(len%k!=0){printf("NO\n");continue;}
k=len/k;
for(i=0;i<len;i+=k)
if(!find(i,i+k-1)){flag=0;break;}
if(flag)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}