#include <cstdio>

constexpr int MAX_N = 1205;
constexpr int MOD = 1000000007;

typedef long long ll;

using namespace std;

int n;
int arr[MAX_N];
ll fact[MAX_N];
ll inv[MAX_N];

ll dp[MAX_N][MAX_N];

int perm(int a, int b) {
    if (b > a || a < 0 || b < 0) return 0;
    if (a == b || b == 0) return fact[b];
    return (int) ((fact[a] * inv[a-b]) % MOD);
}

int pow(int b, int p) {
    if (p == 0) return 1;
    ll ans = pow(b, p >> 1);
    ans = (ans * ans) % MOD;
    if (p & 1) ans = (ans * b) % MOD;
    return (int) ans;
}

int main() {
    scanf(" %d", &n);
    for (int i = 1; i <= n; i++) {
        scanf(" %d", &arr[i]);
    }

    fact[0] = 1;
    for (int i = 1; i <= n; i++) {
        fact[i] = (fact[i-1] * i) % MOD;
        inv[i] = pow(fact[i], MOD - 2);
    }

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= i; j++) {
            if (j != 1 && arr[i-j+1] > arr[i-j+2]) break;
            if (i == j) {
                dp[i][j] = 1;
                break;
            }
            for (int k = j; k <= i; k++) {
                // if (dp[i-j][k] == 0) break;
                // printf("adding %d times %d (i = %d, j = %d, k = %d)\n", (int) dp[i-j][k], perm(k, j), i, j, k);
                dp[i][j] = (dp[i][j] + dp[i-j][k] * perm(k, j)) % MOD;
            }
            // printf("dp[%d][%d] = %d\n", i, j, (int) dp[i][j]);
        }
    }

    ll ans = 0;
    for (int j = 1; j <= n; j++) {
        ans = (ans + dp[n][j]) % MOD;
    }

    printf("%d\n", (int) ans);
    
    return 0;
}