<p>Given <code>n</code> orders, each order consists of a pickup and a delivery service.</p>

<p>Count all valid pickup/delivery possible sequences such that delivery(i) is always after of&nbsp;pickup(i).&nbsp;</p>

<p>Since the answer&nbsp;may be too large,&nbsp;return it modulo&nbsp;10^9 + 7.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 1
<strong>Explanation:</strong> Unique order (P1, D1), Delivery 1 always is after of Pickup 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 6
<strong>Explanation:</strong> All possible orders: 
(P1,P2,D1,D2), (P1,P2,D2,D1), (P1,D1,P2,D2), (P2,P1,D1,D2), (P2,P1,D2,D1) and (P2,D2,P1,D1).
This is an invalid order (P1,D2,P2,D1) because Pickup 2 is after of Delivery 2.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> 90
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 500</code></li>
</ul>

class Solution {
public:
    int mod = 1e9+7;

    vector<long long>fact;

    long long f(int n){
        if(n == 0) return fact[n] = 1;
        return fact[n] = (n * (f(n-1)%mod))%mod;
    }

    long long dp[502][502];

    long long solve(int p,int d, int n){
        if(p == 0) return dp[p][d] = f(d);

        if(dp[p][d] != - 1) return dp[p][d];

        if(p < d) {
            return dp[p][d] = (p*solve(p-1,d,n)%mod + (d-p) * solve(p,d-1,n)%mod)%mod;
        }
        
        return dp[p][d] = p * solve(p-1,d,n)%mod;
    }

    int countOrders(int n) {
        memset(dp,-1,sizeof(dp));
        fact.resize(n+1);
        f(n);
        return solve(n,n,n)%mod;
    }
};
10100