Problem: Let's denote d(n) as the number of divisors of a positive integer n. You are given three integers a, b and c. Your task is to calculate the following sum:

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Find the sum modulo 1073741824 (230).

Input

The first line contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 2000).

Output

Print a single integer — the required sum modulo 1073741824 (230).

Examples

Input

2 2 2


Output

20


Input

4 4 4


Output

328


Input

10 10 10


Output

11536

Note

For the first example.

  * d(1·1·1) = d(1) = 1; 
  * d(1·1·2) = d(2) = 2; 
  * d(1·2·1) = d(2) = 2; 
  * d(1·2·2) = d(4) = 3; 
  * d(2·1·1) = d(2) = 2; 
  * d(2·1·2) = d(4) = 3; 
  * d(2·2·1) = d(4) = 3; 
  * d(2·2·2) = d(8) = 4. 



So the result is 1 + 2 + 2 + 3 + 2 + 3 + 3 + 4 = 20.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool np[2000 + 5] = {1, 1};
int p[2000 + 5];
unordered_map<long long, int> dp;
long long getid(int cnt, int a, int b, int c) {
  int A[4] = {cnt, a, b, c};
  long long res = 0;
  for (int i = 0; i <= 3; i++) res = res * 2000 + A[i];
  return res;
}
void add(int &x, int y) { x = x + y < (1 << 30) ? x + y : x + y - (1 << 30); }
int cal(int cnt, int a, int b, int c) {
  if (a == 0 || b == 0 || c == 0) return 0;
  if (cnt == 0) return 1;
  long long id = getid(cnt, a, b, c);
  if (dp.find(id) != dp.end()) return dp[id];
  int res = 0;
  int x = a;
  for (int i = 0; i <= 13; i++) {
    if (x == 0) break;
    int y = b;
    for (int j = 0; j <= 13; j++) {
      if (y == 0) break;
      int z = c;
      for (int k = 0; k <= 13; k++) {
        if (z == 0) break;
        add(res, 1ll * cal(cnt - 1, x, y, z) * (i + j + k + 1) % (1 << 30));
        z /= p[cnt];
      }
      y /= p[cnt];
    }
    x /= p[cnt];
  }
  dp[id] = res;
  return res;
}
int main() {
  int a, b, c;
  scanf("%d%d%d", &a, &b, &c);
  int n = max(a, max(b, c));
  int num = 0;
  for (int i = 2; i <= n; i++) {
    if (np[i] == 0) p[++num] = i;
    for (int j = 1; j <= num; j++) {
      long long v = 1ll * i * p[j];
      if (v > n) break;
      np[v] = 1;
      if (i % p[j] == 0) break;
    }
  }
  printf("%d\n", cal(num, a, b, c));
  return 0;
}