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;
char a[100000 + 10];
char b[100000 + 10];
char c[100000 + 10];
int vis[100000 + 10];
int main() {
  cin >> n >> t;
  cin >> a;
  cin >> b;
  t = n - t;
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) cnt++;
  }
  if (cnt == t) {
    for (int i = 0; i < n; i++) {
      if (a[i] != b[i]) {
        for (char j = 'a'; j <= 'z'; j++)
          if (a[i] != j && b[i] != j) {
            c[i] = j;
            break;
          }
      } else
        c[i] = a[i];
    }
    printf("%s\n", c);
  } else if (cnt > t) {
    int m = cnt - t;
    for (int i = 0; i < n; i++) {
      if (a[i] != b[i]) {
        for (char j = 'a'; j <= 'z'; j++)
          if (a[i] != j && b[i] != j) {
            c[i] = j;
            break;
          }
      } else {
        c[i] = a[i];
        for (char j = 'a'; j <= 'z'; j++)
          if (a[i] != j && m != 0) {
            c[i] = j;
            m--;
            break;
          }
      }
    }
    printf("%s\n", c);
  } else {
    int m = t - cnt;
    int mm = m;
    int x = n - cnt;
    if (x < m * 2)
      printf("-1\n");
    else {
      for (int i = 0; i < n; i++) {
        if (a[i] == b[i])
          c[i] = a[i];
        else {
          if (m != 0) {
            c[i] = a[i];
            m--;
          } else if (mm != 0) {
            c[i] = b[i];
            mm--;
          } else {
            for (char j = 'a'; j <= 'z'; j++)
              if (a[i] != j && b[i] != j) {
                c[i] = j;
                break;
              }
          }
        }
      }
      printf("%s\n", c);
    }
  }
}