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 a[100010];
char b[100010];
char c[100010];
int d[100010];
int main() {
  int n, t;
  while (cin >> n >> t) {
    memset(a, 0, sizeof(a));
    memset(c, 0, sizeof(c));
    memset(b, 0, sizeof(b));
    memset(d, 0, sizeof(d));
    int y = n - t;
    int ss = 0;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    for (int i = 0; i < n; i++) {
      cin >> b[i];
    }
    for (int i = 0; i < n; i++) {
      if (a[i] == b[i]) {
        ss++;
        d[i] = 1;
      }
    }
    if ((n + 1 - ss) / 2 <= t) {
      if (ss >= y) {
        for (int i = 0; i < n; i++) {
          if (d[i] && y > 0) {
            c[i] = a[i];
            y--;
          } else {
            if ('c' != a[i] && 'c' != b[i]) {
              c[i] = 'c';
            } else if ('b' != a[i] && 'b' != b[i]) {
              c[i] = 'b';
            } else {
              c[i] = 'x';
            }
          }
        }
      } else {
        int k1 = y - ss;
        int k2 = k1;
        for (int i = 0; i < n; i++) {
          if (d[i]) {
            c[i] = a[i];
          } else if (k1 == 0 && k2 == 0) {
            if ('a' != a[i] && 'a' != b[i]) {
              c[i] = 'a';
            } else if ('b' != a[i] && 'b' != b[i]) {
              c[i] = 'b';
            } else {
              c[i] = 'x';
            }
          } else if (k1 > 0) {
            c[i] = a[i];
            k1--;
          } else if (k1 == 0 && k2 > 0) {
            c[i] = b[i];
            k2--;
          }
        }
      }
    } else {
      cout << -1 << endl;
      continue;
    }
    for (int i = 0; i < n; i++) {
      cout << c[i];
    }
    cout << endl;
  }
  return 0;
}