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;
char reqc(char a, char b, int c) {
  if (c == 0) {
    if (a != 'a')
      return a - 1;
    else
      return a + 1;
  } else {
    if (abs(a - b) == 1) {
      if (a == 'a' || b == 'a')
        return 'c';
      else {
        if (a < b)
          return a - 1;
        else
          return b - 1;
      }
    } else
      return (a + b) / 2;
  }
}
int main() {
  int i, j, k, l, m, n, t, sc, dc, occ, bcc, scc, x;
  cin >> n >> t;
  char a[100000], b[100000], ch;
  getchar();
  for (i = 0; i < n; i++) a[i] = getchar();
  getchar();
  for (i = 0; i < n; i++) b[i] = getchar();
  for (i = 0, sc = 0, dc = 0; i < n; i++) {
    if (a[i] != b[i])
      dc++;
    else
      sc++;
  }
  if (t < (dc + 1) / 2)
    printf("-1\n");
  else {
    if (t > dc)
      scc = t - dc;
    else
      scc = 0;
    if (t <= dc) {
      occ = (dc - t) * 2;
      bcc = dc - occ;
    } else {
      bcc = dc;
      occ = 0;
    }
    for (i = 0, x = 1; i < n; i++) {
      if (a[i] != b[i]) {
        if (occ) {
          if (x == 1) {
            printf("%c", a[i]);
            occ--;
            x = 2;
          } else {
            printf("%c", b[i]);
            occ--;
            x = 1;
          }
        } else {
          printf("%c", reqc(a[i], b[i], 1));
        }
      } else {
        if (scc) {
          printf("%c", reqc(a[i], b[i], 0));
          scc--;
        } else
          printf("%c", a[i]);
      }
    }
  }
  return 0;
}