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#pragma GCC optimize ("O3")
#include <bits/stdc++.h>
using namespace std;

#define FOR(i, b, e) for(int i = (b); i < (e); i++)
#define sz(x) int(x.size())
#define all(x) x.begin(), x.end()
#define pb push_back
#define mp make_pair
#define st first
#define nd second
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;

auto &operator<<(auto &o, pair<auto, auto> p) {
	return o << "(" << p.st << ", " << p.nd << ")"; }
auto operator<<(auto &o, auto x)->decltype(end(x), o) {
	o << "{"; int i=0; for(auto e: x) o << ", " + 2*!i++ << e;
	return o << "}"; }
#ifdef LOCAL
#define deb(x...) cerr << "[" #x "]: ", [](auto...$) { \
					((cerr << $ << "; "),...) << endl; }(x)
#else
#define deb(...)
#endif

#include <ext/pb_ds/assoc_container.hpp> /** keep-include */
using namespace __gnu_pbds;
// For places where hacking might be a problem:
const int R = 124'412'612;
struct chash { // To use most bits rather than just the lowest
	const uint64_t C = ll(4e18 * acos(0)) | 71;
	ll operator()(ll x) const { 
		return __builtin_bswap64((x^R)*C); }
};

void solve() {
	int n;
	cin >> n;
	vi ac(n);
	for(int &x: ac) cin >> x;
	vi pref(n + 1);
	FOR(i, 0, n) pref[i + 1] = pref[i] + ac[i];
	ll ans = 0;
	{
		gp_hash_table<int, int, chash> kox({},{},{},{}, {1 << 16});
		FOR(mid_r, 1, n) {
			FOR(first_l, 0, mid_r - 1) FOR(mid_l, 0, mid_r) {
				int sum = pref[mid_r - 1] - pref[first_l] - pref[mid_l];
				kox[sum]++;
			}
			FOR(first_r, 1, mid_r - 1) FOR(first_l, 0, first_r) {
				int sum = -pref[mid_r - 1] + pref[first_r] - pref[first_l];
				kox[sum]++;
			}
			FOR(last_r, mid_r + 1, n + 1) FOR(last_l, 0, last_r) {
				int sum = pref[last_r] - pref[last_l] + pref[mid_r];
				ans += kox[-sum];
			}
		}
	}
	gp_hash_table<int, int, chash> cnt({},{},{},{}, {1 << 16});
	FOR(i, 1, n + 1) FOR(j, 0, i) cnt[pref[i] - pref[j]]++;
	FOR(i, 1, n + 1) {
		FOR(j, 0, i) cnt[pref[i] - pref[j]]--;
		FOR(j, 0, i) {
			FOR(k, 0, j) {
				int sum = pref[i] * 2 - pref[j] - pref[k];
				ans += cnt[-sum];
				cnt[pref[i] - pref[k]]++;
			}
			FOR(k, 0, j) cnt[pref[i] - pref[k]]--;
		}
		FOR(j, 0, i) cnt[pref[i] - pref[j]]++;
	}
	cout << ans << '\n';
}

int main() {
	cin.tie(0)->sync_with_stdio(0);
	int tt = 1;
	// cin >> tt;
	FOR(te, 0, tt) solve();
	return 0;
}