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 = 100 * 1000 + 100;
int n, t;
string s1, s2;
vector<int> v1;
vector<int> v2;
char ans[MaxN];
int ceil(int a, int b) {
  if (a % b == 0) return a / b;
  return (a - a % b) / b + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> t;
  cin >> s1 >> s2;
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i])
      v1.push_back(i);
    else
      v2.push_back(i);
  }
  int x = v1.size();
  if (t < ceil(n - x, 2)) {
    cout << -1 << endl;
    return 0;
  }
  int tmp = max(0, t - (n - x));
  for (int i = 0; i < tmp; i++) {
    if (s1[v1[i]] == 'a')
      ans[v1[i]] = 'b';
    else
      ans[v1[i]] = s1[v1[i]] - 1;
  }
  for (int i = tmp; i < x; i++) ans[v1[i]] = s1[v1[i]];
  t -= tmp;
  int b1 = n - x - t;
  int b2 = b1;
  int b3 = t - b1;
  for (int i = 0; i < b1; i++) ans[v2[i]] = s1[v2[i]];
  for (int i = b1; i < b1 + b2; i++) ans[v2[i]] = s2[v2[i]];
  for (int i = b1 + b2; i < b1 + b2 + b3; i++) {
    set<char> s;
    s.clear();
    s.insert('a');
    s.insert('b');
    s.insert('c');
    s.erase(s1[v2[i]]);
    s.erase(s2[v2[i]]);
    ans[v2[i]] = *s.begin();
  }
  for (int i = 0; i < n; i++) cout << ans[i];
  cout << endl;
  return 0;
}