Juan designed a very unorthodox encryption algorithm. This works as follows:
Given a string $A = a_0a_1a_2 \ldots a_{n-1}$, and a key $K = k_0k_1k_2\ldots k_{m - 1}$, the algorithm encrypts the string $A$, in a resultant string $B = b_0b_1b_2 \ldots b_{n-1}$, where $b_i = a_i + k_{i \mod m}$. We define the sum of two characters $x'$ and $y'$ as the letter that is at position $z'$ in the alphabet, where $z'$ is the result of summing the number of the position of the letter $x'$ in the alphabet plus the number of the position of the letter $y'$ in the alphabet modulo $26$ (cyclically). All positions in the alphabet are indexed starting from $0$.
Example: The encryption of the string $A$ = caribe with the key $K$ = icpc will return the string $B$ = kcgkjg, through the following steps:
- Letter k is at position $10 = (\, 2 \, + \, 8\, ) _{\mod 26} $, c is at position $2$ and i is at position $8$.
- Letter c is at position $ 2\,= (\, 0 \, + \, 2\, )_{ \mod 26}$, a is at position $0$ and c is at position $2$.
- Letter g is at position $ 6\, = ( 17 + 15 )_{\mod 26}$, r is at position $17$ and p is at position $15$.
- Letter k is at position $10 = (\, 8 \, + \, 2\, )_{\mod 26}$, i is at position $8$ and c is at position $2$.
- Letter j is at position $ 9 = (\, 1\, + \, 8\, )_{\mod 26}$, b is at position $1$ and i is at position $8$.
- Letter g is at position $6 = (\, 4 \, + \, 2\, )_{\mod 26}$, e is at position $4$ and c is at position $2$.
Juan started encrypting a text yesterday and today he needs to continue. However, he has lost $K$. Help Juan to find the key, so he can continue working. Juan will give you a string, $A$, and its corresponding encryption, $B$.
Output
Print the key, a string of lowercase letters, if you could find it, or -1, if it is impossible to get $B$ from $A$ with a key of size $m$.