Problem:
Score : 1000 points

Problem Statement

In a two-dimensional plane, we have a rectangle R whose vertices are (0,0), (W,0), (0,H), and (W,H), where W and H are positive integers.
Here, find the number of triangles \Delta in the plane that satisfy all of the following conditions:

Each vertex of \Delta is a grid point, that is, has integer x- and y-coordinates.

\Delta and R shares no vertex.

Each vertex of \Delta lies on the perimeter of R, and all the vertices belong to different sides of R.

\Delta contains at most K grid points strictly within itself (excluding its perimeter and vertices).

Constraints

1 \leq W \leq 10^5

1 \leq H \leq 10^5

0 \leq K \leq 10^5

Input

Input is given from Standard Input in the following format:

W H K

Output

Print the answer.

Sample Input 1

2 3 1

Sample Output 1

12

For example, the triangle with the vertices (1,0), (0,2), and (2,2) contains just one grid point within itself and thus satisfies the condition.

Sample Input 2

5 4 5

Sample Output 2

132

Sample Input 3

100 100 1000

Sample Output 3

461316

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
#include<numeric>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
constexpr ll mod = 1000000009;
const ll INF = mod * mod;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-12;
const ld pi = acos(-1.0);


struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n%mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a*a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 1 << 18;
modint fact[max_n], factinv[max_n];
void init_f() {
	fact[0] = modint(1);
	for (int i = 0; i < max_n - 1; i++) {
		fact[i + 1] = fact[i] * modint(i + 1);
	}
	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
	for (int i = max_n - 2; i >= 0; i--) {
		factinv[i] = factinv[i + 1] * modint(i + 1);
	}
}
modint comb(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[b] * factinv[a - b];
}
modint combP(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[a - b];
}
int gcd(int a, int b) {
	if (a < b)swap(a, b);
	while (b) {
		int r = a % b; a = b; b = r;
	}
	return a;
}


ll calc(int w, int h,int k) {
	if (w < h) {
		int inf = -2 * h - w;
		int res1 = 0; int res2 = 0;
		for (int x1 = 1; x1 < w; x1++) {
			if (2 * k - 2 - x1 * h < inf)break;
			for (int x2 = x1; x2 < w; x2++) {
				if (2 * k - 2 - x1 * h - (x2 - x1) < inf)break;
				for (ll y = 1; y < h; y++) {
					int s = 2 * k - 2 - x1 * h - y * (x2 - x1);
					if (s < inf)break;
					if (s >= 0) {
						if (x1 != x2)res2++;
						else res1++;
					}
					else {
						int ri = gcd(x1, y) + gcd(h - y, x2) + gcd(x2 - x1, h);
						if (s >= -ri) {
							if (x1 != x2)res2++;
							else res1++;
						}
					}
				}
			}
		}
		return 2 * (res1 + 2 * res2);
	}
	else {
		int inf = -2 * h - w;
		int res1 = 0; int res2 = 0;
		for (int x1 = 1; x1 < w; x1++) {
			if (2 * k - 2 - x1 * h < inf)break;
			for (int y = 1; y < h; y++) {
				for (int x2 = x1; x2 < w; x2++) {
					int s = 2 * k - 2 - x1 * h - y * (x2 - x1);
					if (s < inf)break;
					if (s >= 0) {
						if (x1 != x2)res2++;
						else res1++;
					}
					else {
						int ri = gcd(x1, y) + gcd(h - y, x2) + gcd(x2 - x1, h);
						if (s >= -ri) {
							if (x1 != x2)res2++;
							else res1++;
						}
					}

				}
			}
		}
		return 2 * (res1 + 2 * res2);
	}
}
void solve() {
	int w, h, k; cin >> w >> h >> k;
	ll ans = calc(w, h, k) + calc(h, w, k);
	cout << ans << "\n";
}


signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
    //cout << fixed << setprecision(6);
	//init_f();
	//init();
	//int t; cin >> t;rep(i,t)
	solve();
	stop
		return 0;
}