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;
const long long oo = 1e18;
const long long N = 1e6;
const int mod = 1e9 + 7;
const double pi = acos(-1);
double dot(complex<double> a, complex<double> b) {
  return (conj(a) * b).real();
}
double cross(complex<double> a, complex<double> b) {
  return (conj(a) * b).imag();
}
int main() {
  int a, b;
  cin >> a >> b;
  int x = a - b, y = a - b, cnt = 0;
  string s, s1, ans;
  cin >> s >> s1;
  for (int i = 0; i < a; i++) {
    if (s[i] == s1[i]) {
      cnt++;
    }
  }
  if (cnt + ((a - cnt) / 2) < x)
    cout << -1;
  else {
    if (cnt >= x) {
      int f = 0;
      for (int i = 0; i < a; i++) {
        if (s[i] == s1[i] && f < x) {
          ans += s[i];
          f++;
        } else {
          for (char j = 'a'; j <= 'z'; j++) {
            if (j != s[i] && j != s1[i]) {
              ans += j;
              break;
            }
          }
        }
      }
    } else {
      x -= cnt;
      y -= cnt;
      for (int i = 0; i < a; i++) {
        if (s[i] == s1[i])
          ans += s[i];
        else if (x) {
          ans += s[i];
          x--;
        } else if (y) {
          ans += s1[i];
          y--;
        } else {
          for (char j = 'a'; j <= 'z'; j++) {
            if (j != s[i] && j != s1[i]) {
              ans += j;
              break;
            }
          }
        }
      }
    }
    cout << ans;
  }
  return 0;
}