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 maxn = 1e5 + 5, INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n, k;
char a[maxn], b[maxn];
char c[maxn];
int f[maxn];
int main() {
  scanf("%d %d", &n, &k);
  scanf("%s", a + 1);
  scanf("%s", b + 1);
  int m = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] != b[i]) {
      m++;
      f[i] = 1;
    }
  }
  if (m - k > k) {
    printf("-1\n");
    return 0;
  }
  if (k >= m) {
    int cnt = 0;
    for (int i = 1; i <= n; ++i) {
      if (f[i]) {
        c[i] = 'a';
        while (c[i] == a[i] || c[i] == b[i]) c[i]++;
      } else {
        if (cnt < k - m) {
          c[i] = 'a';
          while (c[i] == a[i] || c[i] == b[i]) c[i]++;
          cnt++;
        } else
          c[i] = a[i];
      }
    }
  } else {
    int cnt = 0;
    for (int i = 1; i <= n; ++i) {
      if (f[i]) {
        if (cnt < k) {
          c[i] = b[i];
          cnt++;
        } else {
          c[i] = a[i];
        }
      } else {
        c[i] = a[i];
      }
    }
    cnt = 0;
    for (int i = 1; i <= n; ++i) {
      if (!f[i]) continue;
      if (cnt < 2 * k - m) {
        c[i] = 'a';
        while (c[i] == a[i] || c[i] == b[i]) c[i]++;
        cnt++;
      } else
        break;
    }
  }
  c[n + 1] = '\0';
  printf("%s", c + 1);
  return 0;
}