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;
const long long INF = 1e18 + 42;
template <typename T>
void pop_front(std::vector<T> &vec) {
  assert(!vec.empty());
  vec.erase(vec.begin());
}
bool compPairF(pair<long long, long long> v1, pair<long long, long long> v2) {
  return v1.first < v2.first;
}
bool compPairS(pair<long long, long long> v1, pair<long long, long long> v2) {
  return v1.second < v2.second;
}
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
long long power(long long x, long long y) {
  long long temp;
  if (y == 0) return 1;
  temp = power(x, y / 2);
  if (y % 2 == 0)
    return temp * temp;
  else
    return x * temp * temp;
}
int spf[1000001];
void sieve() {
  spf[1] = 1;
  for (int i = 2; i < 1000001; i++) spf[i] = i;
  for (int i = 4; i < 1000001; i += 2) spf[i] = 2;
  for (int i = 3; i * i < 1000001; i++) {
    if (spf[i] == i) {
      for (int j = i * i; j < 1000001; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
vector<long long> getFactorization(long long x) {
  vector<long long> ret;
  while (x != 1) {
    ret.push_back(spf[x]);
    x = x / spf[x];
  }
  return ret;
}
long long fact(long long n) {
  long long ans = 1;
  for (long long i = 1; i < n + 1; i++) {
    ans = (ans * i) % 1073741824;
  }
  return ans;
}
vector<long long> p;
int q[100];
void pre() {
  for (int i = 2; i < 100; i++) {
    bool flag = true;
    for (int j = 2; j * j <= i; j++) {
      if (i % j == 0) {
        flag = false;
        break;
      }
    }
    if (flag) p.push_back(i);
  }
}
void sol() {
  pre();
  long long A, B, C, i, j, k, ret = 0;
  cin >> A >> B >> C;
  long long ans = 0;
  for (int a = 1; a <= A; a++)
    for (int b = 1; b <= B; b++)
      for (int c = 1; c <= C; c++) {
        int t = a * b * c;
        memset(q, 0, sizeof(q));
        for (int i = 0; i < p.size(); i++)
          while (t % p[i] == 0) q[i]++, t /= p[i];
        int d = 1;
        for (int i = 0; i < p.size(); i++) d = d * (q[i] + 1) % (1 << 30);
        ret = (ret + d) % (1 << 30);
      }
  cout << ret % 1073741824 << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int a = 1;
  sieve();
  while (a--) {
    sol();
  }
}