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;
  string a, b;
  cin >> n >> t;
  cin >> a >> b;
  int eq = 0;
  string ans = "";
  for (int i = 0; i < (n); i++) {
    if (a[i] == b[i]) {
      eq++;
      if (eq <= (n - t))
        ans += a[i];
      else
        ans += "*";
    } else
      ans += "*";
  }
  int eq1 = eq;
  int eq2 = eq;
  if (eq >= (n - t)) {
    for (int i = 0; i < (n); i++) {
      if (ans[i] == '*') {
        for (int j = 0; j < (26); j++) {
          char c = (char)('a' + j);
          if (c != a[i] and c != b[i]) {
            ans[i] = c;
            break;
          }
        }
      }
    }
  } else {
    for (int i = 0; i < (n); i++) {
      if (ans[i] != '*') continue;
      if (eq1 <= eq2) {
        if (eq1 < (n - t)) {
          ans[i] = a[i];
          eq1++;
        }
      } else {
        ans[i] = b[i];
        eq2++;
      }
      if (ans[i] == '*') {
        for (int j = 0; j < (26); j++) {
          char c = (char)('a' + j);
          if (c != a[i] and c != b[i]) {
            ans[i] = c;
            break;
          }
        }
      }
    }
  }
  int dif1 = 0;
  int dif2 = 0;
  for (int i = 0; i < (n); i++) {
    if (ans[i] != a[i]) dif1++;
    if (ans[i] != b[i]) dif2++;
  }
  if (dif1 == t and dif2 == t)
    cout << ans << endl;
  else
    cout << -1 << endl;
  return 0;
}