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, t, ot;
string a, b, c = "";
char other(char a, char b) {
  for (char c = 'a'; c <= 'z'; ++c) {
    if (c != a && c != b) return c;
  }
}
char other(char c) { return 'a' + (c - 'a' + 1) % 26; }
int f(string a, string b) {
  int ret = 0;
  for (int i = 0; i < a.length(); i++) ret += a[i] != b[i];
  return ret;
}
int main() {
  cin >> n >> t >> a >> b;
  ot = t;
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    cnt += a[i] != b[i];
  }
  if (cnt == t) {
    for (int i = 0; i < n; i++) {
      c += (a[i] == b[i]) ? a[i] : other(a[i], b[i]);
    }
    if (ot != f(c, a) || ot != f(c, b))
      puts("-1");
    else
      puts(c.c_str());
  } else if (cnt < t) {
    t = t - cnt;
    for (int i = 0; i < n; i++) {
      if (a[i] == b[i]) {
        if (t != 0) {
          t--;
          c += other(a[i]);
        } else
          c += a[i];
      } else
        c += other(a[i], b[i]);
    }
    if (ot != f(c, a) || ot != f(c, b))
      puts("-1");
    else
      puts(c.c_str());
  } else {
    t = cnt - t;
    t *= 2;
    for (int i = 0; i < n; i++) {
      if (a[i] == b[i])
        c += a[i];
      else {
        if (t > 0) {
          c += (t & 1) ? a[i] : b[i];
          t--;
        } else {
          c += other(a[i], b[i]);
        }
      }
    }
    if (ot != f(c, a) || ot != f(c, b))
      puts("-1");
    else
      puts(c.c_str());
  }
  return 0;
}