#include <bits/stdc++.h>
using namespace std;
int n, t;
char a[1000005], b[1000005], sol[1000005];
vector<int> Egal, Inegal;
inline int Pow_Log(int x, int p) {
  int sol = 1;
  while (p) {
    if (p & 1) {
      sol = (1LL * sol * x) % 1000000007;
      --p;
    }
    x = (1LL * x * x) % 1000000007;
    p >>= 1;
  }
  return sol;
}
inline int GCD(int x, int y) {
  int r = x % y;
  while (r) {
    x = y;
    y = r;
    r = x % y;
  }
  return y;
}
int main() {
  int i, j, k;
  cin.sync_with_stdio(0);
  cin >> n >> t;
  cin >> (a + 1) >> (b + 1);
  for (i = 1; i <= n; ++i)
    if (a[i] == b[i])
      Egal.push_back(i);
    else
      Inegal.push_back(i);
  int p1 = Egal.size(), p2 = Inegal.size();
  for (k = 0; k <= p2; ++k) {
    if ((p2 - k) % 2) continue;
    int x = (p2 - k) / 2, y;
    y = p2 - k - x;
    if (x < 0 || y < 0 || x > p2 || y > p2) continue;
    if (k + x > t || t - k - x > p1) continue;
    int kkt = t - k - x;
    for (i = 0; i < kkt; ++i)
      if (a[Egal[i]] != 'a')
        sol[Egal[i]] = 'a';
      else
        sol[Egal[i]] = 'b';
    for (i = kkt; i < p1; ++i) sol[Egal[i]] = a[Egal[i]];
    for (i = 0; i < k; ++i) {
      for (j = 0; j < 26 && (char('a' + j) == a[Inegal[i]] ||
                             char('a' + j) == b[Inegal[i]]);
           ++j)
        ;
      sol[Inegal[i]] = char('a' + j);
    }
    for (i = 0; i < x; ++i) {
      sol[Inegal[i + k]] = b[Inegal[i + k]];
    }
    for (i = 0; i < x; ++i) {
      sol[Inegal[i + k + x]] = a[Inegal[i + k + x]];
    }
    for (i = 1; i <= n; ++i) cout << sol[i];
    return 0;
  }
  cout << -1;
  return 0;
}

### Prompt

Please formulate 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;
int n, t;
char a[1000005], b[1000005], sol[1000005];
vector<int> Egal, Inegal;
inline int Pow_Log(int x, int p) {
  int sol = 1;
  while (p) {
    if (p & 1) {
      sol = (1LL * sol * x) % 1000000007;
      --p;
    }
    x = (1LL * x * x) % 1000000007;
    p >>= 1;
  }
  return sol;
}
inline int GCD(int x, int y) {
  int r = x % y;
  while (r) {
    x = y;
    y = r;
    r = x % y;
  }
  return y;
}
int main() {
  int i, j, k;
  cin.sync_with_stdio(0);
  cin >> n >> t;
  cin >> (a + 1) >> (b + 1);
  for (i = 1; i <= n; ++i)
    if (a[i] == b[i])
      Egal.push_back(i);
    else
      Inegal.push_back(i);
  int p1 = Egal.size(), p2 = Inegal.size();
  for (k = 0; k <= p2; ++k) {
    if ((p2 - k) % 2) continue;
    int x = (p2 - k) / 2, y;
    y = p2 - k - x;
    if (x < 0 || y < 0 || x > p2 || y > p2) continue;
    if (k + x > t || t - k - x > p1) continue;
    int kkt = t - k - x;
    for (i = 0; i < kkt; ++i)
      if (a[Egal[i]] != 'a')
        sol[Egal[i]] = 'a';
      else
        sol[Egal[i]] = 'b';
    for (i = kkt; i < p1; ++i) sol[Egal[i]] = a[Egal[i]];
    for (i = 0; i < k; ++i) {
      for (j = 0; j < 26 && (char('a' + j) == a[Inegal[i]] ||
                             char('a' + j) == b[Inegal[i]]);
           ++j)
        ;
      sol[Inegal[i]] = char('a' + j);
    }
    for (i = 0; i < x; ++i) {
      sol[Inegal[i + k]] = b[Inegal[i + k]];
    }
    for (i = 0; i < x; ++i) {
      sol[Inegal[i + k + x]] = a[Inegal[i + k + x]];
    }
    for (i = 1; i <= n; ++i) cout << sol[i];
    return 0;
  }
  cout << -1;
  return 0;
}

```