<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>

typedef long long ll;
vector<vector<ll>> dp;
constexpr int N = 500;
constexpr ll MOD = 1e9 + 7;
class Solution {
public:
    int countOrders(int n) {
        if (dp.empty()) {
            dp = vector<vector<ll>>(N + 1, vector<ll>(N + 1));
            dp[0][0] = 1;
            for (int j = 1; j <= N; j++)
                dp[0][j] = dp[0][j - 1] * j % MOD;
            for (int i = 1; i <= N; i++) {
                for (int j = i; j <= N; j++) {
                    ll num = dp[i - 1][j] * i % MOD;
                    ll over = j - i;
                    if (over)
                        num = (num + dp[i][j - 1] * over) % MOD;
                    dp[i][j] = num;
                }
            }
        }
        return dp[n][n];
    }
};
10200