Problem: Marina loves strings of the same length and Vasya loves when there is a third string, different from them in exactly t characters. Help Vasya find at least one such string.

More formally, you are given two strings s1, s2 of length n and number t. Let's denote as f(a, b) the number of characters in which strings a and b are different. Then your task will be to find any string s3 of length n, such that f(s1, s3) = f(s2, s3) = t. If there is no such string, print  - 1.

Input

The first line contains two integers n and t (1 ≤ n ≤ 105, 0 ≤ t ≤ n).

The second line contains string s1 of length n, consisting of lowercase English letters.

The third line contain string s2 of length n, consisting of lowercase English letters.

Output

Print a string of length n, differing from string s1 and from s2 in exactly t characters. Your string should consist only from lowercase English letters. If such string doesn't exist, print -1.

Examples

Input

3 2
abc
xyc


Output

ayd

Input

1 0
c
b


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline void R(int &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
const int maxn = 1e5 + 3;
int cnt;
char solve(char a, char b) {
  if ('b' != a && 'b' != b) return 'b';
  if ('a' != a && 'a' != b) return 'a';
  if ('z' != a && 'z' != b) return 'z';
}
char s1[maxn], s2[maxn];
int main() {
  int n, t;
  R(n);
  R(t);
  t = n - t;
  scanf("%s%s", s1 + 1, s2 + 1);
  for (int i = 1; i <= n; i += 1) {
    cnt += (s1[i] == s2[i]);
  }
  if (t <= cnt) {
    for (int i = 1; i <= n; i += 1) {
      if (s1[i] == s2[i] && t > 0) {
        printf("%c", s1[i]);
        t--;
      } else {
        printf("%c", solve(s1[i], s2[i]));
      }
    }
    return 0;
  } else {
    int temp = t - cnt;
    int pos = 0;
    if (n - cnt < temp * 2) {
      printf("%d\n", -1);
      return 0;
    }
    for (int i = 1; i <= n; i += 1) {
      if (s1[i] != s2[i] && pos < 2) {
        temp--;
        if (!pos)
          printf("%c", s1[i]);
        else
          printf("%c", s2[i]);
        if (!temp) {
          pos += 1;
          temp = t - cnt;
        }
      } else {
        if (s1[i] == s2[i]) {
          printf("%c", s1[i]);
        } else {
          printf("%c", solve(s1[i], s2[i]));
        }
      }
    }
    return 0;
  }
}