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 int INF = 0x7fffffff;
const int maxn = 1e5 + 10;
char a[maxn], b[maxn], c[maxn];
bool f[maxn];
int main() {
  int n, t;
  cin >> n >> t;
  t = n - t;
  cin >> a;
  cin >> b;
  int x = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) {
      f[i] = true;
      x++;
    }
  }
  if (x >= t) {
    int tem = 0;
    for (int i = 0; i < n; i++) {
      if (tem < t) {
        if (f[i]) {
          c[i] = a[i];
          tem++;
        } else {
          for (char j = 'a'; j <= 'z'; j++)
            if (j != a[i] && j != b[i]) {
              c[i] = j;
              break;
            }
        }
      } else {
        for (char j = 'a'; j <= 'z'; j++)
          if (j != a[i] && j != b[i]) {
            c[i] = j;
            break;
          }
      }
    }
  } else {
    if (n >= x + (t - x) * 2) {
      int tem1 = 0, tem2 = 0, need = t - x;
      for (int i = 0; i < n; i++) {
        if (f[i])
          c[i] = a[i];
        else {
          if (tem1 < need) {
            c[i] = a[i];
            tem1++;
          } else if (tem2 < need) {
            c[i] = b[i];
            tem2++;
          } else {
            for (char j = 'a'; j <= 'z'; j++)
              if (j != a[i] && j != b[i]) {
                c[i] = j;
                break;
              }
          }
        }
      }
    } else {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << c << endl;
  return 0;
}