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;
int n, m, i, j, k, l, a[1000009], b[1000009], p[1000009], ans, cn, t, x, y, z,
    mx, mn, s, flag;
char c[1000009], d[1000009], ch[1000009];
int main() {
  while (scanf("%d %d", &n, &t) != EOF) {
    memset(a, 0, sizeof(a));
    flag = 0;
    memset(ch, 0, sizeof(ch));
    x = 0;
    scanf("%s%s", c, d);
    for (i = 0; i < n; i++) {
      if (c[i] == d[i]) {
        x++;
      }
    }
    y = (n - x) / 2;
    ans = x + y;
    t = n - t;
    if (t > ans) {
      printf("-1\n");
    } else {
      z = 0;
      for (i = 0; i < n; i++) {
        if ((z < y) && (z < t) && (c[i] != d[i])) {
          if (flag == 0) {
            flag = 1;
            ch[i] = c[i];
            a[i] = 1;
          } else {
            flag = 0;
            z++;
            ch[i] = d[i];
            a[i] = 1;
          }
        }
      }
      for (i = 0; i < n; i++) {
        if ((z < t) && (c[i] == d[i])) {
          ch[i] = c[i];
          a[i] = 1;
          z++;
        }
      }
      for (i = 0; i < n; i++) {
        if (a[i] == 0) {
          for (j = 'a'; j <= 'z'; j++) {
            if ((j != c[i]) && (j != d[i])) {
              ch[i] = j;
              break;
            }
          }
        }
      }
    }
    printf("%s\n", ch);
  }
  return 0;
}