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 long long input(void) {
  char t;
  long long x = 0;
  int neg = 0;
  t = getchar();
  while ((t < 48 || t > 57) && t != '-') t = getchar();
  if (t == '-') {
    neg = 1;
    t = getchar();
  }
  while (t >= 48 && t <= 57) {
    x = (x << 3) + (x << 1) + t - 48;
    t = getchar();
  }
  if (neg) x = -x;
  return x;
}
inline void output(long long x) {
  char a[20];
  int i = 0, j;
  a[0] = '0';
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x == 0) putchar('0');
  while (x) {
    a[i++] = x % 10 + 48;
    x /= 10;
  }
  for (j = i - 1; j >= 0; j--) {
    putchar(a[j]);
  }
  putchar('\n');
}
int main() {
  long long n, t, diff = 0, i, j, curr, x = 0;
  char s1[100010], s2[100010], ans[100010];
  n = input();
  t = input();
  scanf("%s", s1);
  scanf("%s", s2);
  for (i = 0; i < n; i++)
    if (s1[i] != s2[i]) diff++;
  if (t < (diff + 1) / 2) {
    output(-1);
    return 0;
  }
  ans[n] = '\0';
  if (t >= diff) {
    for (i = 0; i < n; i++) {
      if (s1[i] != s2[i]) {
        for (j = 0; j < 26; j++) {
          if (j + 'a' != s1[i] && j + 'a' != s2[i]) {
            ans[i] = j + 'a';
            break;
          }
        }
      }
    }
    t -= diff;
    for (i = 0; i < n; i++) {
      if (s1[i] == s2[i]) {
        if (t == 0)
          ans[i] = s1[i];
        else {
          if (s1[i] == 'z')
            ans[i] = 'a';
          else
            ans[i] = s1[i] + 1;
          t--;
        }
      }
    }
    printf("%s\n", ans);
    return 0;
  }
  curr = 1;
  for (i = 0; i < n; i++) {
    if (s1[i] != s2[i]) {
      if (curr <= 2 * (diff - t)) {
        if (curr == diff && diff % 2) {
          for (j = 0; j < 26; j++) {
            if (j + 'a' != s1[i] && j + 'a' != s2[i]) {
              ans[i] = j + 'a';
              break;
            }
          }
          x++;
        } else {
          if (curr % 2)
            ans[i] = s1[i];
          else
            ans[i] = s2[i];
          if (curr % 2 == 0) x++;
          curr++;
        }
      } else {
        for (j = 0; j < 26; j++) {
          if (j + 'a' != s1[i] && j + 'a' != s2[i]) {
            ans[i] = j + 'a';
            break;
          }
        }
        x++;
      }
    }
  }
  t -= x;
  for (i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      if (t == 0)
        ans[i] = s1[i];
      else {
        if (s1[i] == 'z')
          ans[i] = 'a';
        else
          ans[i] = s1[i] + 1;
        t--;
      }
    }
  }
  printf("%s\n", ans);
  return 0;
}