B'Xenia is an amateur programmer. Today on the IT lesson she learned about the Hamming distance. The Hamming distance between two strings s xe2 x80 x89= xe2 x80 x89s1s2... sn and t xe2 x80 x89= xe2 x80 x89t1t2... tn of equal length n is value . Record [si xe2 x80 x89 xe2 x89 xa0 xe2 x80 x89ti] is the Iverson notation and represents the following: if si xe2 x80 x89 xe2 x89 xa0 xe2 x80 x89ti, 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, xe2 x80 x89x, xe2 x80 x89m, xe2 x80 x89y. The first line contains two integers n and m (1 xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89n, xe2 x80 x89m xe2 x80 x89 xe2 x89 xa4 xe2 x80 x891012). 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. Print a single integer -- the required Hamming distance. Please, do not use the %lld specifier to read or write 64-bit integers in xd0 xa1++. It is preferred to use the cin, cout streams or the %I64d specifier. 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.'...