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[100005], g[100005], b[100005];
int l[100005];
void solve() {
  for (int i = 0; i < n; ++i) g[i] = a[i];
  for (int i = 1; i <= l[0]; ++i) {
    for (int j = 'a'; j <= 'z'; ++j) {
      char c = j;
      if (c != a[l[i]] && c != b[l[i]]) {
        g[l[i]] = c;
        break;
      }
    }
  }
  int k = t - l[0];
  if (k != 0)
    for (int i = 0; i < n; ++i)
      if (a[i] == b[i]) {
        for (int j = 'a'; j <= 'z'; ++j) {
          char c = j;
          if (c != a[i]) {
            g[i] = c;
            break;
          }
        }
        --k;
        if (!k) break;
      }
  printf("%s\n", g);
}
void init() {
  scanf("%d%d", &n, &t);
  scanf("%s", a);
  scanf("%s", b);
  for (int i = 0; i < n; ++i)
    if (a[i] != b[i]) l[++l[0]] = i;
  if (t >= l[0]) {
    solve();
    return;
  }
  if (l[0] > 2 * t) {
    printf("-1\n");
    return;
  }
  int k = l[0] - t;
  for (int i = 0; i < n; ++i) g[i] = a[i];
  for (int i = k + 1; i <= k * 2; ++i) g[l[i]] = b[l[i]];
  for (int i = k * 2 + 1; i <= l[0]; ++i) {
    for (int j = 'a'; j <= 'z'; ++j) {
      char c = j;
      if (c != a[l[i]] && c != b[l[i]]) {
        g[l[i]] = c;
        break;
      }
    }
  }
  printf("%s\n", g);
}
int main() {
  init();
  return 0;
}