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 main() {
  int n, t;
  cin >> n >> t;
  int same = n - t;
  vector<pair<char, char> > v;
  char a, b;
  string ans = "";
  string s1, s2;
  cin >> s1 >> s2;
  for (int i = 0; i < n; ++i) {
    a = s1[i];
    b = s2[i];
    if (a == b) {
      if (same) {
        v.push_back(make_pair(a, b));
        ans += a;
        same--;
      } else {
        v.push_back(make_pair(a, b));
        ans += " ";
      }
    } else {
      v.push_back(make_pair(a, b));
      ans += " ";
    }
  }
  if (same == 0) {
    for (int i = 0; i < n; ++i) {
      if (ans[i] == ' ') {
        char tem = 'a';
        while (tem == v[i].first || tem == v[i].second) tem++;
        ans[i] = tem;
      }
    }
    cout << ans << endl;
  } else {
    int d1 = same, d2 = same;
    int now = 1;
    for (int i = 0; i < n; ++i) {
      if (ans[i] == ' ') {
        if (d1 == 0 && d2 == 0) {
          char tem = 'a';
          while (tem == v[i].first || tem == v[i].second) tem++;
          ans[i] = tem;
        } else {
          if (now == 1) {
            now = 2;
            d1--;
            ans[i] = v[i].first;
          } else {
            now = 1;
            d2--;
            ans[i] = v[i].second;
          }
        }
      }
    }
    if (d1 == 0 && d2 == 0)
      cout << ans << endl;
    else
      cout << -1 << endl;
  }
}