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 int N = 1e5 + 10;
char s1[N], s2[N], s3[N];
int n, t, diff = 0;
int main() {
  cin >> n >> t;
  scanf("%s", s1 + 1);
  scanf("%s", s2 + 1);
  for (int i = 1, _b = (n); i <= _b; i++) {
    if (s1[i] != s2[i])
      diff++;
    else
      s3[i] = s1[i];
  }
  if (diff > 2 * t) return cout << -1, 0;
  if (diff > t) {
    int equ = diff - t;
    int e1 = equ, e2 = equ;
    for (int i = 1, _b = (n); i <= _b; i++) {
      if (s1[i] != s2[i]) {
        if (e1) {
          e1--;
          s3[i] = s1[i];
        } else if (e2) {
          e2--;
          s3[i] = s2[i];
        } else {
          for (int j = 'a'; j <= 'z'; j++) {
            if (j != s1[i] && j != s2[i]) {
              s3[i] = j;
              break;
            }
          }
        }
      }
    }
  } else {
    int d = t - diff;
    for (int i = 1, _b = (n); i <= _b; i++) {
      if (s1[i] == s2[i]) {
        if (d) {
          for (int j = 'a'; j <= 'z'; j++) {
            if (j != s1[i] && j != s2[i]) {
              s3[i] = j;
              break;
            }
          }
          d--;
        }
      } else {
        for (int j = 'a'; j <= 'z'; j++) {
          if (j != s1[i] && j != s2[i]) {
            s3[i] = j;
            break;
          }
        }
      }
    }
  }
  for (int i = 1, _b = (n); i <= _b; i++) cout << s3[i];
}