//#include <tourist>
#include <bits/stdc++.h>
//#include <atcoder/all>
using namespace std;
//using namespace atcoder;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<ll, ll> p;
const int INF = 1e9;
const ll LINF = ll(1e18);
const int MOD = 1000000007;
const double eps = 1e-7;
const int dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
const int Dx[8] = {0, 1, 1, 1, 0, -1, -1, -1}, Dy[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
#define yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define no cout << "No" << endl
#define NO cout << "NO" << endl
#define rep(i, n) for (int i = 0; i < n; i++)
#define ALL(v) v.begin(), v.end()
#define debug(v)          \
    cout << #v << ":";    \
    for (auto x : v)      \
    {                     \
        cout << x << ' '; \
    }                     \
    cout << endl;
template <class T>
bool chmax(T &a, const T &b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
bool chmin(T &a, const T &b)
{
    if (b < a)
    {
        a = b;
        return 1;
    }
    return 0;
}
//cout<<fixed<<setprecision(15);有効数字15桁
//-std=c++14
//g++ yarudake.cpp -std=c++17 -I .
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
//約数列挙オーダー√n
vector<ll> divisor(ll n)
{
    vector<ll> res;
    for (ll i = 1; i * i <= n; i++)
        if (n % i == 0)
        {
            res.push_back(i);
            if (i != n / i)
                res.push_back(n / i);
        }
    return res;
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n;
    cin >> n;
    n*=n;
    vector<ll> d = divisor(n);
    sort(ALL(d));
    int ans;
    //debug(d);
    
    rep(i, d.size())
    {
        for (int j = i; j < d.size(); j++)
        {
            if(n%(d[i]*d[j]))continue;
            ll b = (d[i]+d[j]);
            ll c = n/(d[i] * d[j]);
            ll t= b*b+4*c;
            if(abs(sqrtl(t)-(ll)sqrtl(t))>eps)continue;
            ll u = (ll)sqrtl(t);
            u-=b;
            if(u<=0||u%2)continue;
            if (d[j] <= u /2&&(d[i]+d[j]+u/2)*u/2*d[j]*d[i]==n)ans++;
            //cout << d[i] << " " << d[j] << " " << u / (2 * d[i] * d[j]) << "\n";
        }
    }
    cout << ans << "\n";
}