#include <bits/stdc++.h>
using namespace std;
vector<int> pset[3];
void initSet(int N, int ind) {
  pset[ind].assign(N, 0);
  for (int i = 0; i < N; i++) pset[ind][i] = i;
}
int findSet(int i, int ind) {
  return (pset[ind][i] == i) ? i : (pset[ind][i] = findSet(pset[ind][i], ind));
}
bool isSameSet(int i, int j, int ind) {
  return findSet(i, ind) == findSet(j, ind);
}
void unionSet(int i, int j, int ind) {
  pset[ind][findSet(i, ind)] = findSet(j, ind);
}
long long powmod(long long base, int exp) {
  if (exp == 1) return base;
  return (base * powmod(base, exp - 1)) % 1000000007;
}
char other(char a, char b) {
  if (a != 'a' && b != 'a') return 'a';
  if (a != 'b' && b != 'b') return 'b';
  return 'c';
}
int main() {
  int n, t;
  cin >> n >> t;
  string a, b;
  cin >> a >> b;
  int diffs = 0;
  vector<int> difinds;
  vector<int> sameinds;
  for (int i = 0; i < n; i++)
    if (a[i] != b[i]) {
      diffs++;
      difinds.push_back(i);
    } else {
      sameinds.push_back(i);
    }
  if (t < (diffs + 1) / 2) {
    cout << -1;
    return 0;
  }
  string res = a;
  for (int i = 0; i < difinds.size() / 2; i++) {
    res[difinds[2 * i]] = a[difinds[2 * i]];
    res[difinds[2 * i + 1]] = b[difinds[2 * i + 1]];
    t--;
  }
  if (diffs % 2 == 1) {
    res[difinds.back()] = other(b[difinds.back()], a[difinds.back()]);
    t--;
  }
  for (int i = 0; i < sameinds.size() && t > 0; i++) {
    res[sameinds[i]] = other(a[sameinds[i]], b[sameinds[i]]);
    t--;
  }
  for (int i = 0; i < difinds.size() / 2 && t > 0; i++) {
    res[difinds[2 * i]] = other(a[difinds[2 * i]], b[difinds[2 * i]]);
    res[difinds[2 * i + 1]] =
        other(a[difinds[2 * i + 1]], b[difinds[2 * i + 1]]);
    t--;
  }
  cout << res;
  cin.get();
  cin.get();
}

### Prompt

Your task is to create a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<int> pset[3];
void initSet(int N, int ind) {
  pset[ind].assign(N, 0);
  for (int i = 0; i < N; i++) pset[ind][i] = i;
}
int findSet(int i, int ind) {
  return (pset[ind][i] == i) ? i : (pset[ind][i] = findSet(pset[ind][i], ind));
}
bool isSameSet(int i, int j, int ind) {
  return findSet(i, ind) == findSet(j, ind);
}
void unionSet(int i, int j, int ind) {
  pset[ind][findSet(i, ind)] = findSet(j, ind);
}
long long powmod(long long base, int exp) {
  if (exp == 1) return base;
  return (base * powmod(base, exp - 1)) % 1000000007;
}
char other(char a, char b) {
  if (a != 'a' && b != 'a') return 'a';
  if (a != 'b' && b != 'b') return 'b';
  return 'c';
}
int main() {
  int n, t;
  cin >> n >> t;
  string a, b;
  cin >> a >> b;
  int diffs = 0;
  vector<int> difinds;
  vector<int> sameinds;
  for (int i = 0; i < n; i++)
    if (a[i] != b[i]) {
      diffs++;
      difinds.push_back(i);
    } else {
      sameinds.push_back(i);
    }
  if (t < (diffs + 1) / 2) {
    cout << -1;
    return 0;
  }
  string res = a;
  for (int i = 0; i < difinds.size() / 2; i++) {
    res[difinds[2 * i]] = a[difinds[2 * i]];
    res[difinds[2 * i + 1]] = b[difinds[2 * i + 1]];
    t--;
  }
  if (diffs % 2 == 1) {
    res[difinds.back()] = other(b[difinds.back()], a[difinds.back()]);
    t--;
  }
  for (int i = 0; i < sameinds.size() && t > 0; i++) {
    res[sameinds[i]] = other(a[sameinds[i]], b[sameinds[i]]);
    t--;
  }
  for (int i = 0; i < difinds.size() / 2 && t > 0; i++) {
    res[difinds[2 * i]] = other(a[difinds[2 * i]], b[difinds[2 * i]]);
    res[difinds[2 * i + 1]] =
        other(a[difinds[2 * i + 1]], b[difinds[2 * i + 1]]);
    t--;
  }
  cout << res;
  cin.get();
  cin.get();
}

```