#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7, MAX = 1e5 + 5;
long long powN(long long a, long long p) {
  if (p == 0) return 1;
  long long z = powN(a, p / 2);
  z = (z * z) % MOD;
  if (p % 2) z = (z * a) % MOD;
  return z;
}
vector<bool> is_prime(MAX + 1, true);
void Sieve() {
  is_prime[0] = is_prime[1] = false;
  int i, j;
  for (i = 2; i * i <= MAX; i++) {
    if (is_prime[i]) {
      for (j = i * i; j <= MAX; j += i) is_prime[j] = false;
    }
  }
}
int main() {
  int n, t;
  cin >> n >> t;
  string s1, s2;
  cin >> s1 >> s2;
  set<int> S;
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i]) S.insert(i);
  }
  if (S.size() < n - t) {
    int x = (n - S.size() + 1) / 2;
    if (x > t)
      cout << -1 << endl;
    else {
      string s3;
      int ganda = 2 * t - n + S.size();
      int done1 = 0, done2 = 0;
      for (int i = 0; i < n; i++) {
        if (S.count(i))
          s3.push_back(s1[i]);
        else {
          if (ganda) {
            for (int j = 0; j < 26; j++) {
              if (s1[i] != 'a' + j && s2[i] != 'a' + j) {
                s3.push_back('a' + j);
                done1++;
                done2++;
                ganda--;
                break;
              }
            }
          } else if (done1 < done2) {
            s3.push_back(s2[i]);
            done1++;
          } else {
            s3.push_back(s1[i]);
            done2++;
          }
        }
      }
      if (done1 == t)
        cout << s3 << endl;
      else {
        int i = 0;
        while (done1 < t && i < n) {
          if (s1[i] == s2[i]) {
            s3[i] = (s1[i] - 'a' + 1) % 26 + 'a';
            done1++;
          }
          i++;
        }
        if (done1 == t)
          cout << s3 << endl;
        else
          cout << -1 << endl;
      }
    }
  } else {
    int done = 0;
    string s3;
    for (int i = 0; i < n; i++) {
      if (S.count(i))
        s3.push_back(s1[i]);
      else {
        for (int j = 0; j < 26; j++) {
          if (s1[i] != 'a' + j && s2[i] != 'a' + j) {
            s3.push_back('a' + j);
            done++;
            break;
          }
        }
      }
    }
    if (done == t)
      cout << s3 << endl;
    else {
      int i = 0;
      while (done < t && i < n) {
        if (s1[i] == s2[i]) {
          s3[i] = (s1[i] - 'a' + 1) % 26 + 'a';
          done++;
        }
        i++;
      }
      if (done == t)
        cout << s3 << endl;
      else
        cout << -1 << endl;
    }
  }
}

### 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, MAX = 1e5 + 5;
long long powN(long long a, long long p) {
  if (p == 0) return 1;
  long long z = powN(a, p / 2);
  z = (z * z) % MOD;
  if (p % 2) z = (z * a) % MOD;
  return z;
}
vector<bool> is_prime(MAX + 1, true);
void Sieve() {
  is_prime[0] = is_prime[1] = false;
  int i, j;
  for (i = 2; i * i <= MAX; i++) {
    if (is_prime[i]) {
      for (j = i * i; j <= MAX; j += i) is_prime[j] = false;
    }
  }
}
int main() {
  int n, t;
  cin >> n >> t;
  string s1, s2;
  cin >> s1 >> s2;
  set<int> S;
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i]) S.insert(i);
  }
  if (S.size() < n - t) {
    int x = (n - S.size() + 1) / 2;
    if (x > t)
      cout << -1 << endl;
    else {
      string s3;
      int ganda = 2 * t - n + S.size();
      int done1 = 0, done2 = 0;
      for (int i = 0; i < n; i++) {
        if (S.count(i))
          s3.push_back(s1[i]);
        else {
          if (ganda) {
            for (int j = 0; j < 26; j++) {
              if (s1[i] != 'a' + j && s2[i] != 'a' + j) {
                s3.push_back('a' + j);
                done1++;
                done2++;
                ganda--;
                break;
              }
            }
          } else if (done1 < done2) {
            s3.push_back(s2[i]);
            done1++;
          } else {
            s3.push_back(s1[i]);
            done2++;
          }
        }
      }
      if (done1 == t)
        cout << s3 << endl;
      else {
        int i = 0;
        while (done1 < t && i < n) {
          if (s1[i] == s2[i]) {
            s3[i] = (s1[i] - 'a' + 1) % 26 + 'a';
            done1++;
          }
          i++;
        }
        if (done1 == t)
          cout << s3 << endl;
        else
          cout << -1 << endl;
      }
    }
  } else {
    int done = 0;
    string s3;
    for (int i = 0; i < n; i++) {
      if (S.count(i))
        s3.push_back(s1[i]);
      else {
        for (int j = 0; j < 26; j++) {
          if (s1[i] != 'a' + j && s2[i] != 'a' + j) {
            s3.push_back('a' + j);
            done++;
            break;
          }
        }
      }
    }
    if (done == t)
      cout << s3 << endl;
    else {
      int i = 0;
      while (done < t && i < n) {
        if (s1[i] == s2[i]) {
          s3[i] = (s1[i] - 'a' + 1) % 26 + 'a';
          done++;
        }
        i++;
      }
      if (done == t)
        cout << s3 << endl;
      else
        cout << -1 << endl;
    }
  }
}

```