#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
const long long MOD = 1e9 + 696969;
const long long INF = 1e18;
int a, b, n, m;
string s1, s2;
int type[1000100];
char rozny(int i) {
  if (s1[i] != 'a' && s2[i] != 'a') return 'a';
  if (s1[i] != 'b' && s2[i] != 'b') return 'b';
  return 'c';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  int la = 0, lb = 0;
  cin >> s1 >> s2;
  int size = s1.size(), dobre = 0;
  for (int i = 0; i < size; ++i)
    if (s1[i] == s2[i]) ++dobre;
  for (int i = 0; i < size; ++i)
    if (s1[i] != s2[i]) {
      if (la < lb)
        type[i] = 1, ++la;
      else
        ++lb, type[i] = 2;
    }
  if (la > m || lb > m) {
    cout << "-1";
    exit(0);
  };
  int br1 = m - la, br2 = m - lb;
  for (int i = 0; i < size; ++i)
    if (s1[i] == s2[i]) {
      if (br1 > 0 && br2 > 0)
        br1--, br2--, ++la, ++lb, type[i] = 3;
      else
        type[i] = -1;
    }
  for (int i = 0; i < size; ++i)
    if (s1[i] != s2[i]) {
      if (type[i] == 1 && lb < m) ++lb, type[i] = 3;
      if (type[i] == 2 && la < m) ++la, type[i] = 3;
    }
  vector<char> wyn;
  wyn.clear();
  for (int i = 0; i < size; ++i) {
    if (type[i] == 1 || type[i] == -1)
      wyn.push_back(s1[i]);
    else if (type[i] == 2)
      wyn.push_back(s2[i]);
    else if (type[i] == 3)
      wyn.push_back(rozny(i));
  }
  int i1 = 0, i2 = 0;
  for (int i = 0; i < wyn.size(); ++i) {
    if (s1[i] != wyn[i]) ++i1;
    if (s2[i] != wyn[i]) ++i2;
  }
  if (i1 != m || i2 != m) {
    cout << "-1";
    exit(0);
  };
  for (int i = 0; i < wyn.size(); ++i) cout << wyn[i];
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
const long long MOD = 1e9 + 696969;
const long long INF = 1e18;
int a, b, n, m;
string s1, s2;
int type[1000100];
char rozny(int i) {
  if (s1[i] != 'a' && s2[i] != 'a') return 'a';
  if (s1[i] != 'b' && s2[i] != 'b') return 'b';
  return 'c';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  int la = 0, lb = 0;
  cin >> s1 >> s2;
  int size = s1.size(), dobre = 0;
  for (int i = 0; i < size; ++i)
    if (s1[i] == s2[i]) ++dobre;
  for (int i = 0; i < size; ++i)
    if (s1[i] != s2[i]) {
      if (la < lb)
        type[i] = 1, ++la;
      else
        ++lb, type[i] = 2;
    }
  if (la > m || lb > m) {
    cout << "-1";
    exit(0);
  };
  int br1 = m - la, br2 = m - lb;
  for (int i = 0; i < size; ++i)
    if (s1[i] == s2[i]) {
      if (br1 > 0 && br2 > 0)
        br1--, br2--, ++la, ++lb, type[i] = 3;
      else
        type[i] = -1;
    }
  for (int i = 0; i < size; ++i)
    if (s1[i] != s2[i]) {
      if (type[i] == 1 && lb < m) ++lb, type[i] = 3;
      if (type[i] == 2 && la < m) ++la, type[i] = 3;
    }
  vector<char> wyn;
  wyn.clear();
  for (int i = 0; i < size; ++i) {
    if (type[i] == 1 || type[i] == -1)
      wyn.push_back(s1[i]);
    else if (type[i] == 2)
      wyn.push_back(s2[i]);
    else if (type[i] == 3)
      wyn.push_back(rozny(i));
  }
  int i1 = 0, i2 = 0;
  for (int i = 0; i < wyn.size(); ++i) {
    if (s1[i] != wyn[i]) ++i1;
    if (s2[i] != wyn[i]) ++i2;
  }
  if (i1 != m || i2 != m) {
    cout << "-1";
    exit(0);
  };
  for (int i = 0; i < wyn.size(); ++i) cout << wyn[i];
}

```