#include <bits/stdc++.h>
using namespace std;
const long long modl = 1000000007;
int farr[1000000];
int pre[1000000];
int cum[1000000];
unsigned long long poow(int base, int po);
void swaap(int *a, int *b);
long long gcd(long long a, long long b);
void mul_big_int(string s1, string s2);
int cnt_prime(int n);
int divisors[2000000];
int main() {
  int a, b, c;
  long long sum = 0;
  int mul;
  cin >> a >> b >> c;
  mul = a * b * c;
  for (int i = 1; i <= mul; i++) {
    for (int j = i; j <= mul; j += i) {
      divisors[j]++;
    }
  }
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j <= b; j++) {
      for (int k = 1; k <= c; k++) {
        sum += divisors[i * j * k];
        sum %= 1073741824;
      }
    }
  }
  cout << sum << "\n";
  return 0;
}
unsigned long long poow(int base, int po) {
  if (po == 0) return 1;
  if (po == 1) return base;
  unsigned long long ans = poow(base, po / 2);
  ans = (ans * ans);
  if (po & 1) return (ans * base);
  return ans;
}
void swaap(int *a, int *b) {
  int temp;
  temp = *a;
  *a = *b;
  *b = temp;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void mul_big_int(string s1, string s2) {
  reverse(s1.begin(), s1.end());
  reverse(s2.begin(), s2.end());
  int mul[1000];
  memset(mul, 0, sizeof(mul));
  long long ln1 = s1.length(), ln2 = s2.length();
  for (int i = 0; i < ln1; i++) {
    for (int j = 0; j < ln2; j++) {
      mul[i + j] += ((s1[i] - '0') * (s2[j] - '0'));
    }
  }
  for (int i = 0; i < 1000; i++) {
    mul[i + 1] += mul[i] / 10;
    mul[i] %= 10;
  }
  int i = 999;
  while (i > 0 && mul[i] == 0) i--;
  for (; i >= 0; i--) {
    cout << mul[i];
  }
  cout << endl;
}
int cnt_prime(int n) {
  vector<bool> prime(n + 1, true);
  prime[0] = false;
  prime[1] = false;
  for (long long i = 2; i * i <= n; i++) {
    if (prime[i]) {
      for (int j = i * i; j <= n; j += i) prime[j] = 0;
    }
  }
  int cntr = 0;
  for (int i = 0; i < (int)prime.size(); i++) {
    if (prime[i]) cntr++;
  }
  return cntr;
}

### Prompt

Your challenge is to write a CPP solution to the following 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:

<image>

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long modl = 1000000007;
int farr[1000000];
int pre[1000000];
int cum[1000000];
unsigned long long poow(int base, int po);
void swaap(int *a, int *b);
long long gcd(long long a, long long b);
void mul_big_int(string s1, string s2);
int cnt_prime(int n);
int divisors[2000000];
int main() {
  int a, b, c;
  long long sum = 0;
  int mul;
  cin >> a >> b >> c;
  mul = a * b * c;
  for (int i = 1; i <= mul; i++) {
    for (int j = i; j <= mul; j += i) {
      divisors[j]++;
    }
  }
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j <= b; j++) {
      for (int k = 1; k <= c; k++) {
        sum += divisors[i * j * k];
        sum %= 1073741824;
      }
    }
  }
  cout << sum << "\n";
  return 0;
}
unsigned long long poow(int base, int po) {
  if (po == 0) return 1;
  if (po == 1) return base;
  unsigned long long ans = poow(base, po / 2);
  ans = (ans * ans);
  if (po & 1) return (ans * base);
  return ans;
}
void swaap(int *a, int *b) {
  int temp;
  temp = *a;
  *a = *b;
  *b = temp;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void mul_big_int(string s1, string s2) {
  reverse(s1.begin(), s1.end());
  reverse(s2.begin(), s2.end());
  int mul[1000];
  memset(mul, 0, sizeof(mul));
  long long ln1 = s1.length(), ln2 = s2.length();
  for (int i = 0; i < ln1; i++) {
    for (int j = 0; j < ln2; j++) {
      mul[i + j] += ((s1[i] - '0') * (s2[j] - '0'));
    }
  }
  for (int i = 0; i < 1000; i++) {
    mul[i + 1] += mul[i] / 10;
    mul[i] %= 10;
  }
  int i = 999;
  while (i > 0 && mul[i] == 0) i--;
  for (; i >= 0; i--) {
    cout << mul[i];
  }
  cout << endl;
}
int cnt_prime(int n) {
  vector<bool> prime(n + 1, true);
  prime[0] = false;
  prime[1] = false;
  for (long long i = 2; i * i <= n; i++) {
    if (prime[i]) {
      for (int j = i * i; j <= n; j += i) prime[j] = 0;
    }
  }
  int cntr = 0;
  for (int i = 0; i < (int)prime.size(); i++) {
    if (prime[i]) cntr++;
  }
  return cntr;
}

```