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;
const long long INF = 1e9;
const long long INFLL = 1e18;
const long long MOD = 1000000007;
const long double eps = 1e-6;
const long double eps2 = 1e-9;
const long long MOD2 = 998244353;
const long long dosz = 5e5;
const long long SZ = (1 << 18);
const long double PI = atan2l(0, -1);
void fast_io() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
long long binpow(long long a, long long n) {
  if (n == 0) return 1;
  if (n % 2 == 1)
    return (binpow(a, n - 1) * a) % MOD;
  else {
    long long b = binpow(a, n / 2);
    return (b * b) % MOD;
  }
}
void solve() {
  long long n, t;
  cin >> n >> t;
  string s1, s2, s3;
  cin >> s1 >> s2;
  s3.resize(n);
  queue<long long> kek;
  for (long long i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      kek.push(i);
    }
  }
  for (long long i = 0; i < n; i++) {
    if (s1[i] != s2[i]) {
      kek.push(i);
    }
  }
  long long cnt1 = 0, cnt2 = 0;
  while (!kek.empty()) {
    long long i = kek.front();
    kek.pop();
    if (s1[i] == s2[i]) {
      if (cnt1 < n - t && cnt2 < n - t) {
        s3[i] = s1[i];
        cnt1++, cnt2++;
      } else {
        if (s1[i] == 'a')
          s3[i] = 'b';
        else
          s3[i] = 'a';
      }
    } else if (s1[i] != s2[i]) {
      if (cnt1 <= cnt2 && cnt1 != n - t) {
        s3[i] = s1[i];
        cnt1++;
      } else if (cnt2 < cnt1 && cnt2 != n - t) {
        s3[i] = s2[i];
        cnt2++;
      } else {
        if (s1[i] != 'c' && s2[i] != 'c') {
          s3[i] = 'c';
        } else if (s1[i] != 'a' && s2[i] != 'a') {
          s3[i] = 'a';
        } else if (s1[i] != 'b' && s2[i] != 'b') {
          s3[i] = 'b';
        }
      }
    }
  }
  if (cnt1 == cnt2 && cnt1 == n - t) {
    cout << s3 << endl;
  } else {
    cout << -1 << endl;
  }
}
signed main() {
  fast_io();
  long long q = 1;
  while (q--) {
    solve();
  }
  return 0;
}