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 N = 1e5 + 10;
const double eps = 1e-9;
int n, t;
char A[N], B[N];
int main() {
  scanf("%d %d\n", &n, &t);
  gets(A);
  gets(B);
  int sameCnt = 0;
  for (int i = 0; i < n; ++i) {
    if (A[i] == B[i]) ++sameCnt;
  }
  int dif = n - sameCnt;
  if (dif / 2 + dif % 2 <= t) {
    int wic = 0;
    int have = 0;
    t -= dif / 2 + dif % 2;
    have = dif % 2;
    if (t > sameCnt) {
      have += 2 * (t - sameCnt);
      t = sameCnt;
    }
    if (n == 1000) {
    }
    for (int i = 0; i < n; ++i) {
      if (have && A[i] != B[i]) {
        char tempval = 'a';
        for (int j = 0; j < 26; ++j) {
          if (tempval + j != A[i] && tempval + j != B[i]) {
            printf("%c", tempval + j);
            break;
          }
        }
        have--;
        continue;
      }
      if (A[i] != B[i]) {
        if (wic)
          printf("%c", A[i]);
        else
          printf("%c", B[i]);
        wic = 1 - wic;
        continue;
      }
      if (t > 0 && A[i] == B[i]) {
        t--;
        char temp = A[i] + 1;
        if (A[i] == 'z') temp = 'y';
        printf("%c", temp);
      } else {
        printf("%c", A[i]);
      }
    }
    printf("\n");
  } else {
    cout << "-1" << endl;
  }
  return 0;
}