B. Xenia and Hamming----数论入门LCA与GCD的应用

B. Xenia and Hamming

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Xenia is an amateur programmer. Today on the IT lesson she learned about the Hamming distance.

The Hamming distance between two strings s = s1s2... sn and t = t1t2... tn of equal length n is value . Record [si ≠ ti] is the Iverson notation and represents the following: if si ≠ ti, it is one, otherwise — zero.

Now Xenia wants to calculate the Hamming distance between two long strings a and b. The first string a is the concatenation of n copies of string x, that is, . The second string b is the concatenation of m copies of string y.

Help Xenia, calculate the required Hamming distance, given n, x, m, y.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 1012). The second line contains a non-empty string x. The third line contains a non-empty string y. Both strings consist of at most 106 lowercase English letters.

It is guaranteed that strings a and b that you obtain from the input have the same length.

Output

Print a single integer — the required Hamming distance.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.

Examples

input

100 10aaaaaaaaaaa

output

0

input

1 1abacabaabzczzz

output

4

input

2 3rzraz

output

5

Note

In the first test case string a is the same as string b and equals 100 letters a. As both strings are equal, the Hamming distance between them is zero.

In the second test case strings a and b differ in their 3-rd, 5-th, 6-th and 7-th characters. Thus, the Hamming distance equals 4.

In the third test case string a is rzrrzr and string b is azazaz. The strings differ in all characters apart for the second one, the Hamming distance between them equals 5.

题目链接：http://codeforces.com/contest/356/problem/B

题目的意思是说给你两个字符串重复的次数，给你两个字符串，问你相同位置不同的字母有多少个。

div1的水题。。

我们先不看重复次数，先看两个字符串的长度，我们求出其lca，总长度是n*len1，那么剩下的len1*n-lca这一些就是在重复lca长度内的不同的次数。

然后我们根据lca开数组，发现有可能开不开，所以我们很自然想到了gcd，这个我们还没有用，我们把字符串分段，分成每段长度为gcd，然后求就可以了。

本以为这个做会出错，MLE等各种错误，但是竟然过了。。。

代码：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define LL long longusing namespace std;char s1[2000000];char s2[2000000];int a[1000010][30];int main(){    LL n,m;    scanf("%I64d%I64d",&n,&m);    scanf("%s%s",s1,s2);    LL len1=strlen(s1);    LL len2=strlen(s2);    LL GCD=__gcd(len1,len2);    LL LCM=len1/GCD*len2;    LL ans=0;    for(int i=0;i<len1;i++){        a[i%GCD][s1[i]-'a']++;    }    for(int i=0;i<len2;i++){        ans+=a[i%GCD][s2[i]-'a'];    }    LL cnt=LCM-ans;    //cout<<LCM<<" "<<ans<<" "<<cnt<<endl;    printf("%I64d\n",n*len1/LCM*cnt);    return 0;}