#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
const int INF = 1 << 30;
const double EPS = 1e-11;
const double PI = acos(-1.0);
void swap(int &a, int &b) {
  int t = a;
  a = b;
  b = t;
}
int max(int a, int b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }
int gcd(int a, int b) {
  while (b) {
    b ^= a ^= b ^= a %= b;
  }
  return a;
}
int main() {
  int t, n, m, i, j;
  char s1[100005], s2[100005], s3[100005];
  int f[100005];
  while (scanf("%d%d", &n, &m) != EOF) {
    memset(f, 0, sizeof(f));
    memset(s3, 0, sizeof(s3));
    scanf("%s", s1);
    scanf("%s", s2);
    t = n - m;
    int num = 0;
    for (i = 0; i < n; i++) {
      if (s1[i] == s2[i]) {
        f[i] = 1;
        num++;
      }
    }
    int dif = n - num;
    int d = dif / 2;
    if (d + num < t) {
      puts("-1");
      continue;
    }
    if (t * 2 - num > n) {
      puts("-1");
      continue;
    }
    int a = 0, b = 0;
    int temp = d;
    if (t > n / 2) {
      temp = min(temp, n / 2);
    } else
      temp = min(t, temp);
    int cnt = t - temp;
    int now = 0;
    int len = 0;
    for (i = 0, j = 0; i < n; i++) {
      if (cnt && f[i]) {
        s3[i] = s1[i];
        cnt--;
        len++;
      } else if (!f[i] && now % 2 == 0 && now < temp * 2) {
        s3[i] = s1[i];
        now++;
        len++;
      } else if (!f[i] && now % 2 == 1 && now < temp * 2) {
        s3[i] = s2[i];
        now++;
        len++;
      }
    }
    for (i = 0; i < n; i++) {
      if (!(s3[i] <= 'z' && s3[i] >= 'a')) {
        for (char s = 'a'; s <= 'z'; ++s) {
          if (s != s1[i] && s != s2[i]) {
            s3[i] = s;
            break;
          }
        }
      }
    }
    s3[n] = '\0';
    puts(s3);
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int maxn = 100005;
const int INF = 1 << 30;
const double EPS = 1e-11;
const double PI = acos(-1.0);
void swap(int &a, int &b) {
  int t = a;
  a = b;
  b = t;
}
int max(int a, int b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }
int gcd(int a, int b) {
  while (b) {
    b ^= a ^= b ^= a %= b;
  }
  return a;
}
int main() {
  int t, n, m, i, j;
  char s1[100005], s2[100005], s3[100005];
  int f[100005];
  while (scanf("%d%d", &n, &m) != EOF) {
    memset(f, 0, sizeof(f));
    memset(s3, 0, sizeof(s3));
    scanf("%s", s1);
    scanf("%s", s2);
    t = n - m;
    int num = 0;
    for (i = 0; i < n; i++) {
      if (s1[i] == s2[i]) {
        f[i] = 1;
        num++;
      }
    }
    int dif = n - num;
    int d = dif / 2;
    if (d + num < t) {
      puts("-1");
      continue;
    }
    if (t * 2 - num > n) {
      puts("-1");
      continue;
    }
    int a = 0, b = 0;
    int temp = d;
    if (t > n / 2) {
      temp = min(temp, n / 2);
    } else
      temp = min(t, temp);
    int cnt = t - temp;
    int now = 0;
    int len = 0;
    for (i = 0, j = 0; i < n; i++) {
      if (cnt && f[i]) {
        s3[i] = s1[i];
        cnt--;
        len++;
      } else if (!f[i] && now % 2 == 0 && now < temp * 2) {
        s3[i] = s1[i];
        now++;
        len++;
      } else if (!f[i] && now % 2 == 1 && now < temp * 2) {
        s3[i] = s2[i];
        now++;
        len++;
      }
    }
    for (i = 0; i < n; i++) {
      if (!(s3[i] <= 'z' && s3[i] >= 'a')) {
        for (char s = 'a'; s <= 'z'; ++s) {
          if (s != s1[i] && s != s2[i]) {
            s3[i] = s;
            break;
          }
        }
      }
    }
    s3[n] = '\0';
    puts(s3);
  }
  return 0;
}

```