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;
void optimizeIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
int main() {
  optimizeIO();
  int n, t;
  cin >> n >> t;
  string s1, s2, s3;
  cin >> s1 >> s2;
  s3 = s1;
  int sa = 0;
  vector<int> v, v2;
  for (typeof(n) i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      sa++;
      v.push_back(i);
    } else
      v2.push_back(i);
  }
  if (n - sa < 2 * (n - t - sa))
    cout << "-1\n";
  else {
    if ((n - t) < sa) {
      int i1 = 0, i2 = 0;
      int dd = n - t;
      for (typeof(n) i = 0; i < n; i++) {
        if (v[i1] == i && dd) {
          s3[i] = s1[i];
          dd--;
          i1++;
        } else {
          if (abs(s1[i] - s2[i]) > 1) {
            s3[i] = min(s1[i], s2[i]) + 1;
          } else {
            if (max(s1[i], s2[i]) == 'z')
              s3[i] = 'a';
            else
              s3[i] = 'z';
          }
        }
      }
    } else {
      int i1 = 0;
      int dd = n - t, c = 0;
      int d1 = n - t - sa, d2 = n - t - sa;
      for (typeof(n) i = 0; i < n; i++) {
        if (i1 < int((v).size())) {
          if (v[i1] == i) {
            s3[i] = s1[i];
            i1++;
            continue;
          }
        }
        if (d1) {
          s3[i] = s1[i];
          d1--;
          continue;
        }
        if (d2) {
          s3[i] = s2[i];
          d2--;
          continue;
        }
        if (d1 == 0 && d2 == 0) {
          if (abs(s1[i] - s2[i]) > 1) {
            s3[i] = min(s1[i], s2[i]) + 1;
          } else {
            if (max(s1[i], s2[i]) == 'z')
              s3[i] = 'a';
            else
              s3[i] = 'z';
          }
        }
      }
    }
    cout << s3 << endl;
  }
  return 0;
}