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 double eps = 1e-9;
const double pi = acos(-1.0);
int n, t;
string s1, s2;
char ans[100100];
int main() {
  cin >> n >> t;
  cin >> s1 >> s2;
  int cur1 = 0, cur2 = 0;
  vector<int> v1, v2;
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i])
      v2.push_back(i);
    else
      v1.push_back(i);
  }
  int n1 = (int)v1.size();
  int n2 = (int)v2.size();
  for (int i = 0; i <= n1; i++) {
    int rem = n1 - i;
    if (rem % 2 != 0) continue;
    if (i + rem / 2 > t) continue;
    int nt = t - i - rem / 2;
    if (nt > n2) continue;
    for (int j = 0; j < rem / 2; j++) {
      ans[v1[j]] = s1[v1[j]];
    }
    for (int j = rem / 2; j < rem; j++) {
      ans[v1[j]] = s2[v1[j]];
    }
    for (int j = rem; j < n1; j++) {
      for (int k = 0; k < 3; k++) {
        char ch = char('a' + k);
        int pos = v1[j];
        if (ch != s1[pos] && ch != s2[pos]) {
          ans[pos] = ch;
          break;
        }
      }
    }
    for (int j = 0; j < nt; j++) {
      if (s1[v2[j]] == 'a')
        ans[v2[j]] = 'b';
      else
        ans[v2[j]] = 'a';
    }
    for (int j = nt; j < n2; j++) {
      ans[v2[j]] = s1[v2[j]];
    }
    for (int j = 0; j < n; j++) {
      printf("%c", ans[j]);
    }
    return 0;
  }
  cout << -1 << endl;
  return 0;
}