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 double eps = 1e-8;
const int maxn = 100005;
char s[2][maxn], ans[maxn];
int n, t;
int cnt;
int main() {
  cin >> n >> t;
  cin >> s[0] >> s[1];
  for (int i = 0; i < n; i++) {
    if (s[0][i] != s[1][i]) cnt++;
  }
  int a, b, c;
  bool flag = 0;
  for (int i = 0; i <= cnt; i++) {
    if ((cnt - i) % 2 == 0) {
      if (i + (cnt - i) / 2 <= t && n - cnt >= t - i - (cnt - i) / 2) {
        flag = 1;
        a = i, b = cnt - i, c = t - i - (cnt - i) / 2;
        break;
      }
    }
  }
  if (!flag)
    cout << -1 << endl;
  else {
    int used = 0;
    int tmp = 0;
    for (int i = 0; i < n; i++) {
      if (s[0][i] != s[1][i]) {
        if (used < b) {
          if (used & 1)
            ans[i] = s[0][i];
          else
            ans[i] = s[1][i];
          used++;
        } else {
          int x = s[0][i] - 'a';
          int y = s[1][i] - 'a';
          if (abs(x - y) > 1)
            ans[i] = 'a' + (x + y) / 2;
          else
            ans[i] = 'a' + (max(x, y) + 1) % 26;
        }
      } else {
        if (tmp < c) {
          ans[i] = 'a' + (s[0][i] - 'a' + 1) % 26;
          tmp++;
        } else
          ans[i] = s[0][i];
      }
    }
    ans[n] = '\0';
    cout << ans << endl;
  }
  return 0;
}