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:

<image>

Find the sum modulo 1073741824 (230).

Input

The first line contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 100).

Output

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

Examples

Input

2 2 2


Output

20


Input

5 6 7


Output

1520

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.
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 110;
const long long mod = 1073741824;
int p[MAXN][100];
bool comp[MAXN];
vector<int> primos;
void crivo() {
  for (int i = 2; i <= 100; i++) {
    if (comp[i] == true) continue;
    if (i * i > 100) break;
    for (int j = i + i; j <= 100; j += i) comp[j] = true;
  }
  for (int i = 2; i <= 100; i++) {
    if (comp[i] == false) primos.push_back(i);
  }
}
int main() {
  memset(comp, false, sizeof comp);
  memset(p, 0, sizeof p);
  int a, b, c;
  cin >> a >> b >> c;
  crivo();
  long long resp = 0;
  map<int, int> m[MAXN];
  for (int i = 1; i <= 100; i++) {
    for (int j = 0; j < primos.size(); j++) {
      if (i % primos[j] == 0) {
        int cont = 0;
        int a = i;
        while (a > 1 && a % primos[j] == 0) a = a / primos[j], cont++;
        m[i][primos[j]] += cont;
      }
    }
  }
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j <= b; j++) {
      for (int k = 1; k <= c; k++) {
        map<int, int> mep;
        long long q = 1;
        for (auto f : m[i]) mep[f.first] += f.second;
        for (auto f : m[j]) mep[f.first] += f.second;
        for (auto f : m[k]) mep[f.first] += f.second;
        for (auto f : mep) q *= f.second + 1;
        resp += q;
        resp = resp % mod;
      }
    }
  }
  cout << resp << endl;
}
