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;
long long readll() {
  bool minus = false;
  long long result = 0;
  char ch;
  ch = getchar();
  while (true) {
    if (ch == '-') break;
    if (ch >= '0' && ch <= '9') break;
    ch = getchar();
  }
  if (ch == '-')
    minus = true;
  else
    result = ch - '0';
  while (true) {
    ch = getchar();
    if (ch < '0' || ch > '9') break;
    result = result * 10 + (ch - '0');
  }
  if (minus)
    return -result;
  else
    return result;
}
int readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int LEFT = 0, RIGHT = 1;
const int MX = 112345;
char s1[MX], s2[MX];
char ans[MX];
vector<int> common, diff;
char getNewChar(int ind) {
  char res = 'a';
  for (int(i) = 0; (i) < (int)(26); (i)++) {
    res += i;
    if (res != s1[ind] && res != s2[ind]) return res;
  }
}
int vis[MX];
int main() {
  memset(vis, 0, sizeof vis);
  int n, t;
  scanf("%d%d", &n, &t);
  scanf("%s", s1);
  scanf("%s", s2);
  for (int(i) = 0; (i) < (int)(n); (i)++) {
    if (s1[i] == s2[i])
      common.push_back(i);
    else
      diff.push_back(i);
  }
  if (t == 0) {
    if (diff.size() > 0)
      puts("-1");
    else
      printf("%s", s1);
    return 0;
  }
  if (diff.size() > t * 2) return puts("-1"), 0;
  int remainDif = diff.size(), remaint = t;
  int cnt = 0;
  while (remainDif > 0 && remainDif > remaint) {
    remainDif -= 2;
    remaint--;
    cnt = diff.back();
    diff.pop_back();
    vis[cnt] = true;
    ans[cnt] = s1[cnt];
    cnt = diff.back();
    diff.pop_back();
    ans[cnt] = s2[cnt];
    vis[cnt] = true;
  }
  while (remaint > 0 && remainDif > 0) {
    remainDif--;
    remaint--;
    cnt = diff.back();
    diff.pop_back();
    vis[cnt] = true;
    ans[cnt] = getNewChar(cnt);
  }
  for (int(i) = 0; (i) < (int)(n); (i)++) {
    if (!vis[i])
      if (remaint > 0) {
        remaint--;
        ans[i] = getNewChar(i);
      } else
        ans[i] = s1[i];
  }
  printf("%s", ans);
}