#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
ll myRand(ll B) { return (ull)rng() % B; }

template <int mod> struct static_modint {
    using mint = static_modint;
    int x;

    static_modint() : x(0) {}
    static_modint(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

    mint &operator+=(const mint &rhs) {
        if ((x += rhs.x) >= mod) x -= mod;
        return *this;
    }
    mint &operator-=(const mint &rhs) {
        if ((x += mod - rhs.x) >= mod) x -= mod;
        return *this;
    }
    mint &operator*=(const mint &rhs) {
        x = (int)(1LL * x * rhs.x % mod);
        return *this;
    }
    mint &operator/=(const mint &rhs) { return *this = *this * rhs.inv(); }

    mint pow(long long n) const {
        mint _x = *this, r = 1;
        while (n) {
            if (n & 1) r *= _x;
            _x *= _x;
            n >>= 1;
        }
        return r;
    }
    mint inv() const { return pow(mod - 2); }

    mint operator+() const { return *this; }
    mint operator-() const { return mint() - *this; }
    friend mint operator+(const mint &lhs, const mint &rhs) { return mint(lhs) += rhs; }
    friend mint operator-(const mint &lhs, const mint &rhs) { return mint(lhs) -= rhs; }
    friend mint operator*(const mint &lhs, const mint &rhs) { return mint(lhs) *= rhs; }
    friend mint operator/(const mint &lhs, const mint &rhs) { return mint(lhs) /= rhs; }
    friend bool operator==(const mint &lhs, const mint &rhs) { return lhs.x == rhs.x; }
    friend bool operator!=(const mint &lhs, const mint &rhs) { return lhs.x != rhs.x; }

    friend ostream &operator<<(ostream &os, const mint &p) { return os << p.x; }
    friend istream &operator>>(istream &is, mint &a) {
        int64_t t;
        is >> t;
        a = static_modint<mod>(t);
        return (is);
    }
};

const unsigned int mod = 1e9 + 7;
using modint = static_modint<mod>;
modint mod_pow(ll n, ll x) { return modint(n).pow(x); }
modint mod_pow(modint n, ll x) { return n.pow(x); }

const int A = 20, B = 75;

// dp[t] = dp[t-B+x] + 3 (0 <= x < i)
// dp[t] = dp[t-B+x] + 2 (i <= x < j)
// dp[t] = dp[t-B+x] + 1 (j <= x < k)
// dp[t] = dp[t-B+x]     (k <= x < B)

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vector<int> t(n);
    for (int i = 0; i < n; ++i) {
        cin >> t[i];
    }
    vector<modint> p2(n, 1);
    for (int i = 1; i < n; ++i) {
        p2[i] = p2[i - 1] * 2;
    }

    vector<vector<vector<modint>>> dp(B + 1, vector<vector<modint>>(B + 1, vector<modint>(B + 1)));
    dp[0][0][0] = 1;
    modint res = 0;
    int pre = 0;
    for (int m = 0; m < n; ++m) {
        int add = t[m] - pre;
        vector<vector<vector<modint>>> ndp(B + 1, vector<vector<modint>>(B + 1, vector<modint>(B + 1)));
        for (int i = 0; i <= B; ++i) {
            for (int j = 0; j <= B; ++j) {
                for (int k = 0; k <= B; ++k) {
                    if (dp[i][j][k] == 0) continue;
                    int a = max(0, i - add);
                    int b = max(0, i - add);
                    int c = max(0, i - add);
                    // 使わない場合
                    ndp[a][b][c] += dp[a][b][c];
                    // 使う場合
                    {
                        vector<int> cost(76);
                        for (int l = 0; l < cost.size(); ++l) {
                            if (l < a) cost[l] = -3;
                            else if (l < b) cost[l] = -2;
                            else if (l < c) cost[l] = -1;
                            else cost[l] = 0;
                        }
                        cost.back() = 1;
                        for (int l = 0; l < cost.size(); ++l) {
                            if (l - A >= 0) cost[l] = min(cost[l], cost[l - A] + 1);
                            if (l - B >= 0) cost[l] = min(cost[l], cost[l - B] + 3);
                        }
                        if (cost.back() >= 1) res += dp[i][j][k] * p2[n - 1 - m] * cost.back();
                        int na = 0, nb = 0, nc = 0;
                        for (int l = 0; l < cost.size(); ++l) {
                            if (cost[l] <= -3) na += 1, nb += 1, nc += 1;
                            else if (cost[l] <= -2) nb += 1, nc += 1;
                            else if (cost[l] <= -1) nc += 1;
                        }
                        dp[na][nb][nc] += dp[a][b][c];
                    }
                }
            }
        }
        pre = t[m];
        swap(dp, ndp);
    }

    cout << res * 2 << endl;
}