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 INF = int(1e9);
const int MOD = int(1e9) + 7;
const unsigned long long N = 1000010;
const long long INFll = 1ll * INF * INF;
const long double ldINF = 1e+018;
const long double EPS = 0.0000001;
const int sdx[4] = {-1, 0, 1, 0};
const int sdy[4] = {0, 1, 0, -1};
const long double PI = 3.1415926535897932384;
template <typename T>
ostream& operator<<(ostream& out, pair<T, T>& a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T>
istream& operator>>(istream& in, pair<T, T>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& a) {
  for (int i = 0; i < a.size(); ++i) in >> a[i];
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> a) {
  for (int i = 0; i < a.size(); ++i)
    if (i == a.size() - 1)
      out << a[i];
    else
      out << a[i] << " ";
  return out;
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, t;
  cin >> n >> t;
  string a, b;
  cin >> a >> b;
  string ans;
  for (int i = 0; i < n; i++) ans += "*";
  int k1 = n - t, k2 = n - t;
  for (int i = 0; i < n && k1 > 0 && k2 > 0; i++) {
    if (k1 > 0 && k2 > 0 && a[i] == b[i]) {
      k1--, k2--;
      ans[i] = a[i];
    }
  }
  for (int i = 0; i < n && (k1 > 0 || k2 > 0); i++) {
    if (ans[i] != '*') continue;
    if (k1 > k2)
      k1--, ans[i] = a[i];
    else
      k2--, ans[i] = b[i];
  }
  if (k1 > 0 || k2 > 0) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (ans[i] != '*') continue;
    for (char j = 'a'; j <= 'z'; j++)
      if (j != a[i] && j != b[i]) {
        ans[i] = j;
        break;
      }
  }
  cout << ans << endl;
  if (0)
    cerr << fixed << setprecision(0)
         << "TIME = " << clock() / (long double)CLOCKS_PER_SEC * 1000 << "\n";
  return 0;
}