<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>

#define ll long long int
class Solution {
public:
ll N=1e9+7;
    ll dp[501][501];
    ll rec(ll curr,ll total,ll n){
        if(total==n && curr==0) return 1;
        ll ans=0;
        if(dp[curr][total]!=-1) return dp[curr][total];
        /*Add*/
        if(total!=n) ans=(ans+rec(curr+1,total+1,n))%N;
        /*Delete*/
        if(curr!=0){
            ll gt=(curr*rec(curr-1,total,n))%N;
            ans=(ans+gt)%N;
        }
        return dp[curr][total]=ans;
    }
    ll fact[501];
    int countOrders(int n) {
        ll gt=1;
        for(ll i=1;i<=500;i++){
            gt=(gt*i)%N;
            fact[i]=gt;
        }
        memset(dp,-1,sizeof(dp));
        ll ans=rec(0,0,n);
        ans=(ans*fact[n])%N;
        return ans;
    }
};
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