#include <algorithm>
#include <iomanip>
#include <array>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <deque>
#include <functional>
#include <iostream>
#include <iterator>
#include <map>
#include <queue>
#include <set>
#include <list>
#include <string>
#include <sstream>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <numeric>
#include <vector>
#include <climits>
using namespace std;
#if __has_include(<atcoder/all>)
#include <atcoder/all>
using namespace atcoder;
#endif
#define GET_MACRO(_1, _2, _3, NAME, ...) NAME
#define _rep(i, n) _rep2(i, 0, n)
#define _rep2(i, a, b) for(int i = (int)(a); i < (int)(b); i++)
#define rep(...) GET_MACRO(__VA_ARGS__, _rep2, _rep)(__VA_ARGS__)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
using i64 = long long;
template<class T, class U>
bool chmin(T& a, const U& b) { return (b < a) ? (a = b, true) : false; }
template<class T, class U>
bool chmax(T& a, const U& b) { return (b > a) ? (a = b, true) : false; }

template<typename T>istream& operator>>(istream&i,vector<T>&v){rep(j,v.size())i>>v[j];return i;}
template<typename T>string join(vector<T>&v){stringstream s;rep(i,v.size())s<<' '<<v[i];return s.str().substr(1);}
template<typename T>ostream& operator<<(ostream&o,vector<T>&v){if(v.size())o<<join(v);return o;}
template<typename T>string join(vector<vector<T>>&vv){string s="\n";rep(i,vv.size())s+=join(vv[i])+"\n";return s;}
template<typename T>ostream& operator<<(ostream&o,vector<vector<T>>&vv){if(vv.size())o<<join(vv);return o;}
template<class T> using pq = priority_queue<T, vector<T>, greater<T>>;

i64 dp[60][6000][60];

int main() {
    i64 n;
    cin >> n;
    vector<i64> a(n);
    cin >> a;

    sort(all(a));
    
    // dp[i][j][k] := i - 1 番目まで見て，総和が j で 選んだ個数が k 個である部分列の数
    dp[0][0][0] = 1;

    i64 ans = 0;

    rep(i, n) {
        rep(j, 5001) {
            rep(k, n) {
                dp[i + 1][j + a[i]][k + 1] += dp[i][j][k];
                dp[i + 1][j][k] += dp[i][j][k];
                if(k) {
                    if((j + a[i])%k == 0 and k * a[i] <= j + a[i]) ans += dp[i][j][k];
                }
            }
        }
    }

    cout << ans << endl;
    

    return 0;
}