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>
char a[100000 + 10];
char b[100000 + 10];
char c[100000 + 10];
int differ[100000 + 10];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  scanf("%s", a);
  scanf("%s", b);
  memset(c, 0, sizeof(c));
  int cou = 0;
  int i;
  for (i = 0; i < n; i++) {
    if (a[i] != b[i]) differ[cou++] = i;
  }
  int tmp = m;
  int now = 0;
  if (m * 2 < cou) {
    printf("-1\n");
    return 0;
  } else if (m >= cou) {
    for (int i = 0; i < cou; i++) {
      for (char j = 'a'; j <= 'z'; j++)
        if (a[differ[i]] != j && b[differ[i]] != j) {
          c[differ[i]] = j;
          break;
        }
    }
    tmp -= cou;
    for (int i = 0; i < n; i++) {
      if (c[i] == 0) {
        if (tmp) {
          for (char j = 'a'; j <= 'z'; j++) {
            if (j != a[i] && j != b[i]) {
              c[i] = j;
              tmp--;
              break;
            }
          }
        } else
          c[i] = a[i];
      }
    }
    for (int i = 0; i < n; i++) {
      printf("%c", c[i]);
    }
    printf("\n");
  } else {
    int A = cou - m;
    if (A * 2 > cou) {
      printf("-1\n");
      return 0;
    }
    int i;
    for (i = 0; i < A; i++) {
      c[differ[i * 2]] = a[differ[i * 2]];
      c[differ[i * 2 + 1]] = b[differ[i * 2 + 1]];
    }
    tmp -= A;
    int now = i * 2;
    for (; tmp > 0; tmp--) {
      if (now < cou) {
        for (char j = 'a'; j <= 'z'; j++) {
          if (j != a[differ[now]] && j != b[differ[now]]) {
            c[differ[now]] = j;
            now++;
            break;
          }
        }
      } else
        break;
    }
    for (int i = 0; i < n; i++) {
      if (c[i] == 0) {
        if (tmp) {
          for (char j = 'a'; j <= 'z'; j++) {
            if (j != a[i] && j != b[i]) {
              c[i] = j;
              tmp--;
              break;
            }
          }
        } else
          c[i] = a[i];
      }
    }
    for (i = 0; i < n; i++) {
      printf("%c", c[i]);
    }
    printf("\n");
  }
  return 0;
}