Problem:
Score : 900 points

Problem Statement

You are developing frog-shaped robots, and decided to race them against each other.

First, you placed N robots onto a number line.
These robots are numbered 1 through N.
The current coordinate of robot i is x_i.
Here, all x_i are integers, and 0 < x_1 < x_2 < ... < x_N.

You will repeatedly perform the following operation:

Select a robot on the number line. Let the coordinate of the robot be x. Select the destination coordinate, either x-1 or x-2, that is not occupied by another robot. The robot now jumps to the selected coordinate.

When the coordinate of a robot becomes 0 or less, the robot is considered finished and will be removed from the number line immediately.
You will repeat the operation until all the robots finish the race.

Depending on your choice in the operation, the N robots can finish the race in different orders.
In how many different orders can the N robots finish the race?
Find the answer modulo 10^9+7.

Constraints

2 ≤ N ≤ 10^5

x_i is an integer.

0 < x_1 < x_2 < ... < x_N ≤ 10^9

Partial Score

In a test set worth 500 points, N ≤ 8.

Input

The input is given from Standard Input in the following format:

N
x_1 x_2 ... x_N

Output

Print the number of the different orders in which the N robots can finish the race, modulo 10^9+7.

Sample Input 1

3
1 2 3

Sample Output 1

4

There are four different orders in which the three robots can finish the race:

(Robot 1 → Robot 2 → Robot 3)

(Robot 1 → Robot 3 → Robot 2)

(Robot 2 → Robot 1 → Robot 3)

(Robot 2 → Robot 3 → Robot 1)

Sample Input 2

3
2 3 4

Sample Output 2

6

There are six different orders in which the three robots can finish the race:

(Robot 1 → Robot 2 → Robot 3)

(Robot 1 → Robot 3 → Robot 2)

(Robot 2 → Robot 1 → Robot 3)

(Robot 2 → Robot 3 → Robot 1)

(Robot 3 → Robot 1 → Robot 2)

(Robot 3 → Robot 2 → Robot 1)

For example, the order (Robot 3 → Robot 2 → Robot 1) can be achieved as shown in the figure below:

Sample Input 3

8
1 2 3 5 7 11 13 17

Sample Output 3

10080

Sample Input 4

13
4 6 8 9 10 12 14 15 16 18 20 21 22

Sample Output 4

311014372

Remember to print the answer modulo 10^9+7.
This case is not included in the test set for the partial score.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#ifdef LOCAL
//#define _GLIBCXX_DEBUG
#endif
//#pragma GCC target("avx512f,avx512dq,avx512cd,avx512bw,avx512vl")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<int, int> Pi;
typedef vector<ll> Vec;
typedef vector<int> Vi;
typedef vector<string> Vs;
typedef vector<char> Vc;
typedef vector<P> VP;
typedef vector<vector<ll>> VV;
typedef vector<vector<int>> VVi;
typedef vector<vector<char>> VVc;
typedef vector<vector<vector<ll>>> VVV;
typedef vector<vector<vector<vector<ll>>>> VVVV;

#define REP(i, a, b) for(ll i=(a); i<(b); i++)
#define PER(i, a, b) for(ll i=(a); i>=(b); i--)
#define rep(i, n) REP(i, 0, n)
#define per(i, n) PER(i, n, 0)
const ll INF=1e18+18;
const ll MOD=1000000007;
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl;
#define YES(n) cout << ((n) ? "YES" : "NO") << endl;
#define ALL(v) v.begin(), v.end()
#define rALL(v) v.rbegin(), v.rend()
#define pb(x) push_back(x)
#define mp(a, b) make_pair(a,b)
#define Each(a,b) for(auto &a :b)
#define rEach(i, mp) for (auto i = mp.rbegin(); i != mp.rend(); ++i)
#ifdef LOCAL
#define dbg(x_) cerr << #x_ << ":" << x_ << endl;
#define dbgmap(mp) cerr << #mp << ":"<<endl; for (auto i = mp.begin(); i != mp.end(); ++i) { cerr << i->first <<":"<<i->second << endl;}
#define dbgset(st) cerr << #st << ":"<<endl; for (auto i = st.begin(); i != st.end(); ++i) { cerr << *i <<" ";}cerr<<endl;
#define dbgarr(n,m,arr) rep(i,n){rep(j,m){cerr<<arr[i][j]<<" ";}cerr<<endl;}
#define dbgdp(n,arr) rep(i,n){cerr<<arr[i]<<" ";}cerr<<endl;
#else
#define dbg(...)
#define dbgmap(...)
#define dbgset(...)
#define dbgarr(...)
#define dbgdp(...)
#endif
#define out(a) cout<<a<<endl
#define out2(a,b) cout<<a<<" "<<b<<endl
#define vout(v) rep(i,v.size()){cout<<v[i]<<" ";}cout<<endl
#define Uniq(v) v.erase(unique(v.begin(), v.end()), v.end())
#define fi first
#define se second

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }

template<typename T1, typename T2>
ostream &operator<<(ostream &s, const pair<T1, T2> &p) { return s<<"("<<p.first<<", "<<p.second<<")"; }

template<typename T>istream& operator>>(istream&i,vector<T>&v)
{rep(j,v.size())i>>v[j];return i;}

// vector
template<typename T>
ostream &operator<<(ostream &s, const vector<T> &v) {
    int len=v.size();
    for(int i=0; i<len; ++i) {
        s<<v[i];
        if(i<len-1) s<<"	";
    }
    return s;
}

// 2 dimentional vector
template<typename T>
ostream &operator<<(ostream &s, const vector<vector<T> > &vv) {
    int len=vv.size();
    for(int i=0; i<len; ++i) {
        s<<vv[i]<<endl;
    }
    return s;
}

//mint
struct mint {
    ll x; // typedef long long ll;
    mint(ll x=0):x((x%MOD+MOD)%MOD){}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint a) {
        if ((x += a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += MOD-a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator*=(const mint a) { (x *= a.x) %= MOD; return *this;}
    mint operator+(const mint a) const { return mint(*this) += a;}
    mint operator-(const mint a) const { return mint(*this) -= a;}
    mint operator*(const mint a) const { return mint(*this) *= a;}
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }

    // for prime MOD
    mint inv() const { return pow(MOD-2);}
    mint& operator/=(const mint a) { return *this *= a.inv();}
    mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
typedef vector<mint> Vmint;
typedef vector<vector<mint>> VVmint;
typedef vector<vector<vector<mint>>> VVVmint;

struct combination {
    vector<mint> fact, ifact;
    combination(int n):fact(n+1),ifact(n+1) {
        assert(n < MOD);
        fact[0] = 1;
        for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
        ifact[n] = fact[n].inv();
        for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
    }
    mint operator()(int n, int k) {
        if (k < 0 || k > n) return 0;
        return fact[n]*ifact[k]*ifact[n-k];
    }
};

int solve(){
    ll n;
    cin>>n;
    Vec a(n);
    cin>>a;
    mint ans = 1;
    ll cnt = 0;
    rep(i,n){
        if(a[i] < cnt*2+1){
            ans *= cnt+1;
            cnt--;
        }
        cnt++;
    }
    REP(i,1,cnt+1){
        ans *= i;
    }
    out(ans);
    return  0;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout<<std::setprecision(10);
    solve();
}