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 <image>. 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, <image>. 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 %I64d specifier.

Examples

Input

100 10
a
aaaaaaaaaa


Output

0


Input

1 1
abacaba
abzczzz


Output

4


Input

2 3
rzr
az


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.
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(0.0) * 2.0;
const double eps = 1e-12;
const int step[8][2] = {{1, 0}, {-1, 0}, {0, 1},  {0, -1},
                        {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
template <class T>
inline T abs1(T a) {
  return a < 0 ? -a : a;
}
template <class T>
inline T max1(T a, T b, T c = -1234567, T d = -1234567) {
  T ans = a > b ? a : b;
  if (c != -1234567) ans = max1(ans, c);
  if (d != -1234567) ans = max1(ans, d);
  return ans;
}
template <class T>
inline T min1(T a, T b, T c = -1234567, T d = -1234567) {
  T ans = a < b ? a : b;
  if (c != -1234567) ans = min(ans, c);
  if (d != -1234567) ans = min(ans, d);
  return ans;
}
template <class T>
inline T gcd1(T a, T b) {
  if (a < b) swap(a, b);
  if (a % b == 0) return b;
  return gcd1(b, a % b);
}
template <class T>
inline T lb(T num) {
  return num & (-num);
}
inline int jud(double a, double b) {
  if (abs(a) < eps && abs(b) < eps)
    return 0;
  else if (abs1(a - b) / abs1(a) < eps)
    return 0;
  if (a < b) return -1;
  return 1;
}
template <typename t>
inline int jud(t a, t b) {
  if (a < b) return -1;
  if (a == b) return 0;
  return 1;
}
template <typename it, typename t1>
inline int find(t1 val, it a, int na, bool f_small = 1, bool f_lb = 1) {
  int be = 0, en = na - 1;
  if (*a <= *(a + na - 1)) {
    if (f_lb == 0)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != 1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != -1)
          en = mid;
        else
          be = mid + 1;
      }
    if (f_small && jud(*(a + be), val) == 1) be--;
    if (!f_small && jud(*(a + be), val) == -1) be++;
  } else {
    if (f_lb)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != -1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != 1)
          en = mid;
        else
          be = mid + 1;
      }
    if (!f_small && jud(*(a + be), val) == -1) be--;
    if (f_small && jud(*(a + be), val) == 1) be++;
  }
  return be;
}
inline int bitnum(unsigned long long nValue) {
  nValue = ((0xaaaaaaaaaaaaaaaaull & nValue) >> 1) +
           (0x5555555555555555ull & nValue);
  nValue = ((0xccccccccccccccccull & nValue) >> 2) +
           (0x3333333333333333ull & nValue);
  nValue = ((0xf0f0f0f0f0f0f0f0ull & nValue) >> 4) +
           (0x0f0f0f0f0f0f0f0full & nValue);
  nValue = ((0xff00ff00ff00ff00ull & nValue) >> 8) +
           (0x00ff00ff00ff00ffull & nValue);
  nValue = ((0xffff0000ffff0000ull & nValue) >> 16) +
           (0x0000ffff0000ffffull & nValue);
  nValue = ((0xffffffff00000000ull & nValue) >> 32) +
           (0x00000000ffffffffull & nValue);
  return nValue;
}
long long pow(long long n, long long m, long long mod = 0) {
  long long ans = 1;
  long long k = n;
  while (m) {
    if (m & 1) {
      ans *= k;
      if (mod) ans %= mod;
    }
    k *= k;
    if (mod) k %= mod;
    m >>= 1;
  }
  return ans;
}
const int maxn = 1001000;
int cnt[maxn][2][30];
long long n, m;
char stra[maxn * 2], strb[maxn * 2];
long long lstra, lstrb;
int main() {
  ios_base::sync_with_stdio(0);
  scanf("%I64d%I64d", &n, &m);
  scanf("%s%s", stra, strb);
  lstra = strlen(stra), lstrb = strlen(strb);
  long long gcd = gcd1(lstra, lstrb);
  for (int i = 0; i < lstra; i++) cnt[i % gcd][0][stra[i] - 'a']++;
  for (int i = 0; i < lstrb; i++) cnt[i % gcd][1][strb[i] - 'a']++;
  long long ans = 0;
  for (int i = 0; i < gcd; i++) {
    for (int j = 0; j < 26; j++)
      ans += cnt[i][0][j] * ((lstrb / gcd) - cnt[i][1][j]);
  }
  ans *= n / (lstrb / gcd);
  while (n / (lstrb / gcd) != m / (lstra / gcd))
    ;
  cout << ans << endl;
  return 0;
}
