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;
void RI() {}
template <typename... T>
void RI(int& head, T&... tail) {
  scanf("%d", &head);
  RI(tail...);
}
const int N = 1e5 + 10;
int n, m;
char a[N], b[N], c[N];
char f(char x, char y) {
  for (int i = ('a'); i <= int('z'); i++)
    if (x != i && y != i) return i;
  assert(0);
}
int main() {
  RI(n, m);
  scanf("%s%s", a, b);
  int da = 0, db = 0;
  for (int i = 0; i < int(n); i++) {
    if (a[i] == b[i])
      c[i] = a[i];
    else {
      if (da < db)
        c[i] = b[i], da++;
      else
        c[i] = a[i], db++;
    }
  }
  if (da > m || db > m) {
    puts("-1");
    return 0;
  }
  for (int i = 0; i < int(n); i++)
    if (da < m && db < m && a[i] == b[i]) {
      c[i] = f(a[i], b[i]);
      da++;
      db++;
    }
  for (int i = 0; i < int(n); i++)
    if (da < m && c[i] == a[i] && c[i] != b[i]) {
      c[i] = f(a[i], b[i]);
      da++;
    }
  for (int i = 0; i < int(n); i++)
    if (db < m && c[i] == b[i] && c[i] != a[i]) {
      c[i] = f(a[i], b[i]);
      db++;
    }
  if (da == m && db == m)
    puts(c);
  else
    puts("-1");
  return 0;
}