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;
const int MAXX = 1e5 + 2;
char a[MAXX];
char b[MAXX];
char c[MAXX];
int n, t, ile, ktory;
int zm[MAXX];
char roz(int x) {
  for (int i = 1; i <= 26; i++)
    if (a[x] != 'a' + i - 1 && b[x] != 'a' + i - 1) return 'a' + i - 1;
}
int main() {
  scanf("%d%d", &n, &t);
  scanf(" %s", a + 1);
  scanf(" %s", b + 1);
  for (int i = 1; i <= n; i++) {
    c[i] = a[i];
    if (a[i] != b[i]) ile++;
  }
  if (t * 2 < ile) {
    printf("-1");
    return 0;
  }
  if (ile % 2 == 1) {
    for (int i = 1; i <= n; i++)
      if (a[i] != b[i]) {
        c[i] = roz(i), t--, zm[i] = 3;
        break;
      }
  }
  for (int i = 1; i <= n; i++)
    if (t && a[i] != b[i] && !zm[i]) {
      if (ktory % 2 == 0)
        c[i] = a[i], zm[i] = 1, ktory++;
      else
        c[i] = b[i], zm[i] = 1, t--, ktory++;
    }
  for (int i = 1; i <= n; i++) {
    if (!t) break;
    if (a[i] == b[i]) c[i] = roz(i), t--;
  }
  int g;
  for (int i = 1; i <= n; i++) {
    if (!t) break;
    if (a[i] != b[i] && zm[i] == 1) {
      c[i] = roz(i), g++;
      if (g % 2 == 0) t--;
    }
  }
  for (int i = 1; i <= n; i++) printf("%c", c[i]);
}