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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline bool read(T &x) {
  int c = getchar();
  int sgn = 1;
  while (~c && c<'0' | c> '9') {
    if (c == '-') sgn = -1;
    c = getchar();
  }
  for (x = 0; ~c && '0' <= c && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  x *= sgn;
  return ~c;
}
struct debugger {
  template <typename T>
  debugger &operator,(const T &v) {
    cerr << v << "    ";
    return *this;
  }
} dbg;
template <class T>
void __stl_print__(T &x) {
  cerr << "[";
  for (__typeof((x).end()) i = (x).begin(); i != (x).end(); ++i)
    cerr << (i != x.begin() ? ", " : "") << *i;
  cerr << "]" << endl;
}
template <class T, class U>
inline T max(T &a, U &b) {
  return a > b ? a : b;
}
template <class T, class U>
inline T min(T &a, U &b) {
  return a < b ? a : b;
}
template <class T, class U>
inline T swap(T &a, U &b) {
  T tmp = a;
  a = b;
  b = tmp;
}
const long long INF = (1ll) << 50;
const int mx = 3e5 + 7;
const int mod = 1000000007;
const double pi = 2 * acos(0.0);
int EQ(double d) {
  if (fabs(d) < 1e-7) return 0;
  return d > 1e-7 ? 1 : -1;
}
char x[mx], y[mx], m[mx];
char c(char u, char v) {
  for (__typeof((26) - 1) i = (0); i <= ((26) - 1); ++i)
    if (u != char(i + 'a') && v != char(i + 'a')) return char(i + 'a');
}
int main() {
  int n = ({
    int a;
    read(a);
    a;
  }),
      t = ({
        int a;
        read(a);
        a;
      });
  scanf("%s %s", x, y);
  int cnt = 0;
  for (__typeof((n)-1) i = (0); i <= ((n)-1); ++i) cnt += (x[i] != y[i]);
  if (cnt <= t) {
    for (__typeof((n)-1) i = (0); i <= ((n)-1); ++i) {
      if (x[i] != y[i]) {
        m[i] = c(x[i], y[i]);
        t--;
      } else
        m[i] = x[i];
    }
    int i = 0;
    while (t) {
      if (x[i] == y[i]) {
        t--;
        m[i] = c(m[i], m[i]);
      }
      i++;
    }
    m[n] = 0;
    puts(m);
  } else {
    int o, p;
    o = cnt - t;
    p = o;
    for (__typeof((n)-1) i = (0); i <= ((n)-1); ++i) {
      if (x[i] != y[i]) {
        if (o) {
          m[i] = x[i];
          o--;
          t--;
        } else if (t) {
          m[i] = c(x[i], y[i]);
          t--;
        } else {
          m[i] = y[i];
          p--;
        }
      } else
        m[i] = x[i];
    }
    if (o == 0 && t == 0 && p == 0) {
      m[n] = 0;
      puts(m);
    } else
      puts("-1");
  }
}