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//Sylwia Sapkowska
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;

void __print(int x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << "'" << x << "'";}
void __print(const char *x) {cerr << '"' << x << '"';}
void __print(const string &x) {cerr << '"' << x << '"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifdef LOCAL
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

#define int long long
typedef pair<int, int> T;
const int A = 3e7;
int tab[2*A+7];

void solve(){
    int n; cin >> n;
    vector<int>a(n+1), pref(n+1);
    for (int i = 1; i<=n; i++) {
        cin >> a[i];
        pref[i] = pref[i-1] + a[i];
    }
    int ans = 0;
    for (int l2 = n; l2 >= 1; l2--){ 
        for (int l1 = n; l1 >= 1; l1--){
            for (int r1 = n; r1 >= l1; r1--){
                tab[pref[r1]-pref[l1-1]+pref[l2]+A]++;
            }
        } 
        for (int l1 = n; l1 >= 1; l1--){ 
            for (int r1 = n; r1 >= l1; r1--){
                int sum = pref[r1] - pref[l1-1] - pref[l2-1];
                ans += tab[-sum+A];
            }
        }
    }
    debug(ans);
    //odjac gdy pewne dwa sa rowne
    //pierwsze dwa rowne, ten pozniejszy wiekszy
    memset(tab, 0, sizeof(tab));
    for (int l1 = n; l1 >= 1; l1--){
        for (int r1 = n; r1 >= l1; r1--){
            ans -= 3*tab[-2*(pref[r1]-pref[l1-1])+A];
            tab[pref[r1]-pref[l1-1]+A]++;
        }
    }
    debug(ans);
    memset(tab, 0, sizeof(tab));
    for (int l1 = 1; l1 <= n; l1++){
        for (int r1 = l1; r1 <= n; r1++){
            ans -= 3*tab[-2*(pref[r1]-pref[l1-1])+A];
            tab[pref[r1]-pref[l1-1]+A]++;
        }
    }
    //wszystkie trzy rowne(?) --> czyli kazdy ma sume 0
    for (int l = 1; l <= n; l++){
        for (int r = l; r <= n; r++){
            if (pref[r] == pref[l-1]){
                debug(l, r);
                ans--;
            }
        }
    }
    cout << ans/6 << "\n";
}

int32_t main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    solve();

    return 0;
}