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 main() {
  int n, t;
  scanf("%d %d", &n, &t);
  char s1[100005], s2[100005];
  scanf("%s", s1);
  scanf("%s", s2);
  char s3[100005];
  int ans = 0;
  vector<int> v;
  int i;
  for (i = 0; i < n; ++i)
    if (s1[i] == s2[i])
      s3[i] = s1[i];
    else
      v.push_back(i);
  if (v.size() > t) {
    int m = 2 * t - v.size();
    if (m < 0)
      ans = -1;
    else {
      for (i = 0; i < m; ++i) {
        int newch1 = s1[v[i]] - 'a' + 1;
        int newch2 = s2[v[i]] - 'a' + 1;
        newch1 = newch1 % 26;
        newch2 = newch2 % 26;
        if (newch1 + 'a' != s2[v[i]])
          s3[v[i]] = 'a' + newch1;
        else
          s3[v[i]] = 'a' + newch2;
      }
      int j = 0;
      m = v.size() - m;
      m = m / 2;
      for (j = 0; j < m; ++i, ++j) s3[v[i]] = s1[v[i]];
      for (j = 0; j < m; ++i, ++j) s3[v[i]] = s2[v[i]];
    }
  } else {
    for (i = 0; i < v.size(); ++i) {
      int newch1 = s1[v[i]] - 'a' + 1;
      int newch2 = s2[v[i]] - 'a' + 1;
      newch1 = newch1 % 26;
      newch2 = newch2 % 26;
      if (newch1 + 'a' != s2[v[i]])
        s3[v[i]] = 'a' + newch1;
      else
        s3[v[i]] = 'a' + newch2;
    }
    int rem = t - v.size();
    for (i = 0; i < n && rem != 0; ++i)
      if (s1[i] == s2[i]) {
        int newchar = s1[i] - 'a' + 1;
        newchar = newchar % 26;
        s3[i] = 'a' + newchar;
        rem--;
      }
  }
  s3[n] = '\0';
  if (ans == -1)
    printf("-1");
  else
    printf("%s", s3);
  return 0;
}