Codeforces Round #324 (Div. 2) C. Marina and Vasya Set

C. Marina and Vasya

Marina loves strings of the same length and Vasya loves when there was a third string, different from them in exactly C3>t characters. Help Vasya find at least one such string.

More formally, given and stringss1, s2 of length n and number T. Let's denote as F(a, b) The number of characters in which strings a and c13>b is different. Then your task'll be the find any string s3 of the length n, such that f (s1, s3) = F(s2, s3) = T. If There is no such string, print -1.

Input

The first line contains integers n and t (1≤ n ≤105, 0≤ t ≤ n).

The second line contains string s1 of length n, consisting of lowercase 中文版 lette Rs.

The third line contain string s2 of length n, consisting of lowercase 中文版 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 中文版 letters. If such string doesn ' t exist, print-1.

Sample Test (s) input

3 2
Abc
Xyc

Output

Ayd

Input

1 0
C
B

Output

-1

Test instructions: give you two strings of the same length as n, so that you can construct a string that is different from any of the two strings that are given with the number of letters T
The puzzle: I am set disorderly, the code ability is poor, beg for understanding

///1085422276#include <bits/stdc++.h>using namespacestd; typedefLong Longll;#defineMem (a) memset (A,0,sizeof (a))#defineMeminf (a) memset (A,127,sizeof (a));#defineTS printf ("111111\n");#definefor (I,A,B) for (int i=a;i<=b;i++)#defineForj (I,A,B) for (int i=a;i>=b;i--)#defineREAD (a,b,c) scanf ("%d%d%d", &a,&b,&c)#defineINF 100000inline ll read () {ll x=0, f=1; CharCh=GetChar ();  while(ch<'0'|| Ch>'9')    {        if(ch=='-') f=-1; CH=GetChar (); }     while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; CH=GetChar (); }    returnx*F;}//****************************************#defineMAXN 100000+10Set<int>G;Set<int>:: iterator it;CharA[MAXN],B[MAXN],C[MAXN];intMain () {intn=read (); intt=read (); scanf ("%s%s", A, b); For (I,0, N-1)     if(a[i]==B[i])         G.insert (i); T=n-T; if(t>g.size ()) {        intTmp= (N-g.size ())/2; if(Tmp<t-g.size ()) {cout<<-1<<endl;return 0;}  for(It=g.begin (); It!=g.end (); it++) c[*it]=a[*it]; T=t-g.size (); intflag=1;intFlag2=0;  for(intI=0; i<n;i++)        {            if(G.count (i) = =0&&flag&&t) {C[i]=A[i]; Flag=0; Flag2=1; }            Else if(G.count (i) = =0&&flag2&&t) {C[i]=B[i]; T--; Flag=1; Flag2=0; }            Else if(G.count (i) = =0){                  for(intj=0;j< -; j + +)                    if('a'+j!=a[i]&&'a'+j!=B[i]) {C[i]='a'+j; Break; }            }        }    }    Else {         for(It=g.begin (); It!=g.end (); it++)        {            if(t!=0) {c[*it]=a[*it]; T--; }Else {              for(intj=0;j< -; j + +)                    if('a'+j!=a[*it]&&'a'+j!=b[*it]) {c[*it]='a'+j; Break; }}        }         for(intI=0; i<n;i++)        {            if(G.count (i) = =0)            {                 for(intj=0;j< -; j + +)                    if('a'+j!=a[i]&&'a'+j!=B[i]) {C[i]='a'+j; Break; }            }        }    }  for(intI=0; i<n;i++) printf ("%c", C[i]); return 0;}

Code

Codeforces Round #324 (Div. 2) C. Marina and Vasya Set