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

#define TRACE(x) x
#define WATCH(x) TRACE(cout << #x" = " << x << endl)
#define WATCHR(a, b) TRACE(for (auto it=a; it!=b;) cout << *(it++) << " "; cout << endl)
#define WATCHC(V) TRACE({cout << #V" = "; WATCHR(V.begin(), V.end());})

#define all(x) (x).begin(), (x).end()

typedef long long ll;
typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vll;
typedef vector<vll> vvll;

const int MOD = 1e9 + 7;
int sum(int a, int b) { return (a + b) % MOD; }
int prod(int a, int b) { return a * 1ll * b % MOD; }

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cout << fixed << setprecision(15);

    const int MAXN = 1500;
    
    vi fact(MAXN);
    fact[0] = 1;
    for (int i = 1; i < MAXN; i++)
        fact[i] = prod(i, fact[i-1]);
    
    vvi ncr(MAXN, vi(MAXN + 1));
    ncr[0][0] = 1;
    for (int n = 1; n < MAXN; n++) {
        ncr[n][0] = 1;
        for (int r = 0; r <= n; r++)
            ncr[n][r] = sum(ncr[n-1][r], ncr[n-1][r-1]);
    }
    
    int N;
    cin >> N;
    
    vi vals(N);
    for (int i = 0; i < N; i++)
        cin >> vals[i];
    
    vvi ways(N + 1, vi(N + 1, 0)); // ways[i][j] = # of ways to arrange vals[i:] into V of size j
    //vvi sums(N + 1, vi(N + 1, 0)); // sums[i][j] = # of ways to arrange vals[i:] into V of size <= j

    ways[N][0] = 1;
    //fill(all(sums[N]), 1);
    
    for (int i = N - 1; i >= 0; i--) {
        for (int j = 1; j <= N; j++) {
            if (i + j > N) break;
            if (j > 1 && vals[i+j-1] < vals[i+j-2]) break;            
            
            for (int k = 0; k <= j; k++) {
                int put = prod(ncr[j][k], fact[k]);
                ways[i][j] = sum(ways[i][j], prod(put, ways[i+j][k]));
            }
        }
        
        /*for (int j = 1; j <= N; j++) {
            sums[i][j] = sum(ways[i][j], sums[i][j-1]);
        }*/
    }
    
    int ans = 0;
    for (int j = 1; j <= N; j++) {
        ans = sum(ans, ways[0][j]);
    }           
    cout << ans << endl;
    return 0;
}