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

using ll = long long;
using VI = vector<int>;
using P = pair<int, int>;

#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define FOR(i, a, b) for (ll i = a; i < (ll)(b); i++)
#define ALL(a) (a).begin(),(a).end()

constexpr int INF = 1001001001;
constexpr ll LINF = 1001001001001001001ll;
constexpr int DX[] = {1, 0, -1, 0};
constexpr int DY[] = {0, 1, 0, -1};

template< typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) {return a < b && (a = b, true); }
template< typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) {return a > b && (a = b, true); }

const ll MOD = 1000000007;

int main() {
    int N;
    cin >> N;
    VI A(N), B(0);
    set<int> st;
    st.insert(0);
    REP(i, N) {
        cin >> A[i];
        st.insert(A[i]);
    }
    map<int, int> mp;
    for (int i : st) {
        mp[i] = B.size();
        B.push_back(i);
    }
    vector<vector<vector<vector<ll>>>> dp(2, vector<vector<vector<ll>>>(51, vector<vector<ll>>(5001, vector<ll>(B.size(), 0))));
    //ll dp[51][51][5001][101];
    //REP(i, 51) REP(j, 51) REP(k, 5001) REP(l, 101) dp[i][j][k][l] = 0;
    // dp[i][j][k][l] i個目まででj個選んで合計がkで最大がA[l]の場合の数
    dp[0][0][0][mp[0]] = 1;
    REP(i, N) {
        //cout << i << endl;
        REP(j, i + 2) {
            REP(k, min(5001, j * 100 + 101)) {
                REP(l, B.size()) {
                    dp[(i + 1) % 2][j][k][l] = 0;
                }
            }
        }
        REP(j, i + 1) {
            REP(k, j * 100 + 1) {
                REP(l, B.size()) {
                    dp[(i + 1) % 2][j][k][l] += dp[i % 2][j][k][l];
                    int x = l;
                    if (A[i] > B[l]) x = mp[A[i]];
                    dp[(i + 1) % 2][j + 1][k + A[i]][x] += dp[i % 2][j][k][l];
                    //if (dp[i % 2][j][k][l] > 0) cout << i << " " << j << " " << k + A[i] << " " << x << " " << dp[(i + 1) % 2][j + 1][k + A[i]][x] << endl;
                }
            }
        }
    }
    ll ans = 0;
        FOR(j, 2, N + 1) {
            REP(k, j * 100 + 1) {
                REP(l, B.size()) {
                    if (k % (j - 1)) continue;
                    if (B[l] * (j - 1) > k) continue;
                    ans += dp[N % 2][j][k][l];
                    //if (dp[N % 2][j][k][l] > 0) cout << j << " " << k << " " << l << " " << dp[N % 2][j][k][l] << endl;
                }
            }
        }
    cout << ans << endl;
}