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

The second line contains integer *k* (1 ≤ *k* ≤ 1000).
string = input()
partskitnekarnehai = int(input())
if len(string)%partskitnekarnehai!=0:
    print('NO')
    exit()
    
nayalength = len(string) // partskitnekarnehai

# o is length of part
for j in range(partskitnekarnehai):
    seedha_part = string[ j*nayalength : (j + 1)*nayalength]
    # print(seedha_part)
    ulta_part=reversed(seedha_part)
    # print("".join(ulta_part))
    if seedha_part == "".join(ulta_part):
        print("YES") 
        exit()
print("NO")

'''
https://codeforces.com/problemset/problem/548/A
Aaj exit() seekha, kaafi useful, i always thought agar aise kuch 
karna hai to i'll have to define a function, fir use return to get
out of the code.
Bas run hojaye time se code, like TLE na aajaye
Are bhai yaaar wo sabse upar wala block mai bhool hi gaya tha,
gahdam ek aur wrong submission
Also line 7 me agar -1 kia to it only check first part of string
'''