#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;

ll cnt(ll X, ll Y, ll R, ll lim) {
    ll a = 0;
    ll b = 10000;
    ll res = 0;

    for (ll i = (Y + R) - (Y + R) % 10000; i >= lim; i -= 10000) {
        while ((b - X) * (b - X) + (i - Y) * (i - Y) <= R * R) {
            b += 10000;
        }
        while ((a - X) * (a - X) + (i - Y) * (i - Y) <= R * R) {
            a -= 10000;
        }

        res += (b - a - 10000) / 10000;
    }
    return res;
}

int main() {
    ld x, y, r;
    cin >> x >> y >> r;

    ll ans = 0;
    ll X = round(x * 10000);
    ll Y = round(y * 10000);
    ll R = round(r * 10000);
    X %= 10000;
    Y %= 10000;

    ans += cnt(X, Y, R, 10000);
    ans += cnt(X, -Y, R, 0);

    cout << ans << endl;

    return 0;
}
#include <bits/stdc++.h>
using namespace std;

int main() {
    int l,r;
    cin>>l>>r;
    long long ans = 0;
    vector<long long> dp(r+1,0);
    for(int g = r; g>=2;g--){
        long long cnt = (r/g - (l-1)/g);
        dp[g] = (cnt * (cnt-1))/2;
        //Currently dp[g] contains g,2g,3g,4g,... as gcd 
        //we need to remove all other gcd's except g
        for(int j=2*g;j<=r;j+=g){
            dp[g] -= dp[j]; //remove gcd = j from dp[g]
        }
        ans += dp[g];
    }
    //dp[g] contains the number of pairs which have gcd exactly g
    //But we need to remove the pairs which have one of number = g
    //there are r/g-1 such pairs
    //Since r/g-1 is the total number of elements which are paired in g
    //Hence we remove them from final answer
    
    //Let's say l < 2 : Here l = 1, and we want gcd!=1, Hence we ignore this
    //l >= 2: Assume l = 4, Now we need not remove r/2-1 , r/3-1 from our answer because such pairs are never formed
    //....as we are considering only pairs from l to r.
    for(int i=max(2,l);i<=r;i++) ans -= r/i - 1;
    cout<<2*ans<<endl;
}