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>
#pragma comment(linker, "/stack:16777216")
using namespace std;
const int INF = 1000000000;
const int MAX = 200024;
const int MAX2 = 310000;
const int MAXD = 20;
const int ADD = 200000;
const int BASE = 1000000007;
const int MOD = 1000000007;
int n, t;
char A[MAX];
char B[MAX];
char C[MAX];
int main() {
  scanf("%d %d", &n, &t);
  scanf("%s %s", A, B);
  int a = 0, b = 0;
  for (int i = (0); i < (n); i++)
    if (A[i] == B[i])
      ++a;
    else
      ++b;
  int cb = -1;
  int ca = -1;
  for (int x = (0); x < (b + 1); x++) {
    if ((b - x) % 2 == 1) continue;
    int w = x;
    w += (b - x) / 2;
    if (w > t) continue;
    if ((t - w) > a) continue;
    cb = x;
    ca = (t - w);
  }
  int p = 0;
  if (ca == -1) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = (0); i < (n); i++) {
    if (A[i] == B[i]) {
      if (ca > 0) {
        for (int c = ('a'); c < ('z' + 1); c++)
          if (c != A[i]) C[i] = c;
        --ca;
      } else {
        C[i] = A[i];
      }
    } else {
      if (cb > 0) {
        for (int c = ('a'); c < ('z' + 1); c++)
          if (c != A[i] && c != B[i]) C[i] = c;
        --cb;
      } else {
        if (p % 2 == 0) {
          for (int c = ('a'); c < ('z' + 1); c++)
            if (c == A[i] && c != B[i]) C[i] = c;
        } else {
          for (int c = ('a'); c < ('z' + 1); c++)
            if (c != A[i] && c == B[i]) C[i] = c;
        }
        ++p;
      }
    }
  }
  for (int i = (0); i < (n); i++) cout << C[i];
  cout << endl;
  return 0;
}