dp of size n + 1 to store the number of derangements for each number from 0 to n.dp[0] = 1 and dp[1] = 0.n, and for each i, calculate dp[i] using the recurrence relation: dp[i] = (i - 1) * (dp[i - 1] + dp[i - 2]).dp[i] modulo 1000000007 to prevent integer overflow.dp[n] as the final answer.