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 double pi = acos(-1.0);
const double pii = 2 * pi;
const double eps = 1e-6;
const long long MOD = 1e9 + 7;
void solve() {
  int n, t;
  cin >> n >> t;
  string x, y;
  cin >> x >> y;
  int ctr = 0;
  for (int i = 0; i < n; i++)
    if (x[i] != y[i]) ctr++;
  if (t > ctr) {
    t -= ctr;
    string ans = x;
    for (int i = 0; i < n; i++) {
      if (x[i] != y[i]) {
        char xx = 'a';
        while (xx == x[i] || xx == y[i]) xx++;
        ans[i] = xx;
      } else {
        if (t > 0) {
          char xx = 'a';
          while (xx == x[i]) xx++;
          ans[i] = xx;
          t--;
        } else {
          ans[i] = x[i];
        }
      }
    }
    cout << ans << "\n";
    return;
  } else {
    int a = 0;
    for (a = 0; a < ctr; a++) {
      if (2 * a <= ctr && ctr - a == t) {
        break;
      }
    }
    if (!(2 * a <= ctr && ctr - a == t)) {
      cout << "-1\n";
      return;
    }
    a *= 2;
    string ans = x;
    for (int i = 0; i < n; i++) {
      if (x[i] == y[i])
        ans[i] = x[i];
      else {
        if (a > 0) {
          if (a % 2 == 0) {
            ans[i] = x[i];
            a--;
          } else {
            ans[i] = y[i];
            a--;
          }
        } else {
          char xx = 'a';
          while (x[i] == xx || y[i] == xx) xx++;
          ans[i] = xx;
        }
      }
    }
    cout << ans << "\n";
    return;
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  solve();
  return 0;
}