#include <bits/stdc++.h>
const double PI = acos(-1.0);
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
inline T Min(T a, T b) {
  return a < b ? a : b;
}
template <class T>
inline T Max(T a, T b) {
  return a > b ? a : b;
}
using namespace std;
int n, m, L;
char ch1[100010];
char ch2[100010];
char ch[100010];
int vis[100010];
int main() {
  int t;
  while (scanf("%d%d", &n, &t) != EOF) {
    scanf("%s", ch1);
    scanf("%s", ch2);
    int sum = 0;
    vector<int> pos;
    memset(vis, 0, sizeof vis);
    for (int i = 0; i < n; i++) {
      if (ch1[i] != ch2[i])
        sum++;
      else
        pos.push_back(i);
    }
    int sm = n - t;
    sm *= 2;
    int size = (int)pos.size();
    for (int i = 0; i < size && sm; i++) {
      int ps = pos[i];
      ch[ps] = ch1[ps];
      sm -= 2;
      vis[ps] = 1;
    }
    int flag = 0;
    for (int i = 0; i < n; i++) {
      if (vis[i]) continue;
      if (sm) {
        if (!flag)
          ch[i] = ch1[i];
        else
          ch[i] = ch2[i];
        sm--;
        flag = 1 - flag;
      } else {
        for (char c = 'a'; c <= 'z'; c++) {
          if (ch1[i] != c && ch2[i] != c) {
            ch[i] = c;
            break;
          }
        }
      }
    }
    ch[n] = '\0';
    if (sm == 0)
      printf("%s\n", ch);
    else
      printf("-1\n");
  }
  return 0;
}

### 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>
const double PI = acos(-1.0);
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
inline T Min(T a, T b) {
  return a < b ? a : b;
}
template <class T>
inline T Max(T a, T b) {
  return a > b ? a : b;
}
using namespace std;
int n, m, L;
char ch1[100010];
char ch2[100010];
char ch[100010];
int vis[100010];
int main() {
  int t;
  while (scanf("%d%d", &n, &t) != EOF) {
    scanf("%s", ch1);
    scanf("%s", ch2);
    int sum = 0;
    vector<int> pos;
    memset(vis, 0, sizeof vis);
    for (int i = 0; i < n; i++) {
      if (ch1[i] != ch2[i])
        sum++;
      else
        pos.push_back(i);
    }
    int sm = n - t;
    sm *= 2;
    int size = (int)pos.size();
    for (int i = 0; i < size && sm; i++) {
      int ps = pos[i];
      ch[ps] = ch1[ps];
      sm -= 2;
      vis[ps] = 1;
    }
    int flag = 0;
    for (int i = 0; i < n; i++) {
      if (vis[i]) continue;
      if (sm) {
        if (!flag)
          ch[i] = ch1[i];
        else
          ch[i] = ch2[i];
        sm--;
        flag = 1 - flag;
      } else {
        for (char c = 'a'; c <= 'z'; c++) {
          if (ch1[i] != c && ch2[i] != c) {
            ch[i] = c;
            break;
          }
        }
      }
    }
    ch[n] = '\0';
    if (sm == 0)
      printf("%s\n", ch);
    else
      printf("-1\n");
  }
  return 0;
}

```