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 n, k;
char s1[100005], s2[100005];
char kq[100005];
void solve() {
  int khac, trug, i, j, dem1, dem2, dem, can, pt;
  pt = 0;
  khac = trug = 0;
  cin >> s1 >> s2;
  for (i = 0; i < n; i++) {
    if (s1[i] != s2[i])
      khac++;
    else
      trug++;
  }
  dem = 0;
  if (khac <= k) {
    can = abs(k - khac);
    for (i = 0; i < n; i++) {
      if (s1[i] == s2[i]) {
        if (dem < can) {
          dem++;
          for (j = 'a'; j <= 'z'; j++) {
            if (j != s1[i]) {
              printf("%c", j);
              break;
            }
          }
        } else
          printf("%c", s1[i]);
        continue;
      }
      for (j = 'a'; j <= 'z'; j++) {
        if (j != s1[i] && j != s2[i]) {
          printf("%c", j);
          break;
        }
      }
    }
  } else {
    pt = 0;
    for (i = 0; i < n; i++) {
      if (s1[i] == s2[i]) {
        kq[pt++] = s1[i];
        continue;
      }
      if (khac > k) {
        dem++;
        if (dem == 2) khac--, dem = 0;
        dem % 2 != 0 ? kq[pt++] = s1[i] : kq[pt++] = s2[i];
      } else {
        for (j = 'a'; j <= 'z'; j++) {
          if (j != s1[i] && j != s2[i]) {
            kq[pt++] = j;
            break;
          }
        }
      }
    }
    dem1 = dem2 = 0;
    kq[pt] = 0;
    for (i = 0; i < pt; i++) {
      if (kq[i] != s1[i]) dem1++;
      if (kq[i] != s2[i]) dem2++;
    }
    if (dem1 == dem2 && dem1 == k)
      cout << kq << endl;
    else
      cout << -1 << endl;
  }
}
int main() {
  while (cin >> n >> k) solve();
  return 0;
}