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;
char a[101000], b[101000];
char ans[101000];
int vis[101000];
char dif(char x, char y) {
  if (x == y) {
    if (x == 'a') return 'b';
    return 'a';
  } else {
    if (x == 'a' || y == 'a') {
      if (x == 'b' || y == 'b') return 'c';
      return 'b';
    }
    return 'a';
  }
}
int main() {
  int i, j, k;
  int n, t;
  int d = 0, s = 0;
  scanf("%d%d", &n, &t);
  scanf("%s%s", a, b);
  for (i = 0; i < n; i++) {
    if (a[i] == b[i])
      s++;
    else
      d++;
  }
  int req = n - t;
  int avl = (d / 2) + s;
  if (avl >= req) {
    for (i = 0; i < n; i++) vis[i] = 0;
    int c1 = req, c2 = req;
    for (i = 0; i < n; i++) {
      if (c1 != 0 && c2 != 0 && a[i] == b[i]) {
        vis[i] = 1;
        ans[i] = a[i];
        c1--;
        c2--;
      }
    }
    for (i = 0; i < n; i++) {
      if (!vis[i]) {
        if (c1 != 0) {
          ans[i] = a[i];
          c1--;
        } else if (c2 != 0) {
          ans[i] = b[i];
          c2--;
        } else {
          ans[i] = dif(a[i], b[i]);
        }
      }
    }
    for (i = 0; i < n; i++) printf("%c", ans[i]);
    printf("\n");
  } else {
    printf("-1\n");
    return 0;
  }
  return 0;
}