A chess knight can move as indicated in the chess diagram below:

 .           

This time, we place our chess knight on any numbered key of a phone pad (indicated above), and the knight makes N-1 hops.  Each hop must be from one key to another numbered key.

Each time it lands on a key (including the initial placement of the knight), it presses the number of that key, pressing N digits total.

How many distinct numbers can you dial in this manner?

Since the answer may be large, output the answer modulo 10^9 + 7.

Example 1:

Input: 1
Output: 10

Example 2:

Input: 2
Output: 20

Example 3:

Input: 3
Output: 46

Note:

• 1 <= N <= 5000
  

國際象棋中的騎士可以按下圖所示進行移動：

 .           

這一次，我們將 “騎士” 放在電話撥號盤的任意數字鍵（如上圖所示）上，接下來，騎士將會跳 N-1 步。每一步必須是從一個數字鍵跳到另一個數字鍵。

每當它落在一個鍵上（包括騎士的初始位置），都會撥出鍵所對應的數字，總共按下 N 位數字。

你能用這種方式撥出多少個不同的號碼？

因為答案可能很大，所以輸出答案模 10^9 + 7。

示例 1：

輸入：1
輸出：10

示例 2：

輸入：2
輸出：20

示例 3：

輸入：3
輸出：46

提示：

• 1 <= N <= 5000

 1 class Solution {
 2     func knightDialer(_ N: Int) -> Int {
 3         var N = N
 4         var mod:Int64 = 1000000007
 5         N -= 1
 6         var dp:[Int64] = [Int64](repeating: 1,count: 10)
 7         for i in 0..<N
 8         {
 9             var ndp:[Int64] = [Int64](repeating: 0,count: 10)
10             ndp[0] = dp[4] + dp[6]
11             ndp[1] = dp[6] + dp[8]
12             ndp[2] = dp[7] + dp[9]
13             ndp[3] = dp[4] + dp[8]
14             ndp[4] = dp[3] + dp[9] + dp[0]
15             ndp[6] = dp[1] + dp[7] + dp[0]
16             ndp[8] = dp[1] + dp[3]
17             ndp[7] = dp[2] + dp[6]
18             ndp[9] = dp[2] + dp[4]
19             for j in 0..<10
20             {
21                 ndp[j] %= mod
22             }
23             dp = ndp
24         }
25         var ret:Int64 = 0
26         for i in 0..<10
27         {
28             ret += dp[i]
29         }
30         return Int(ret % mod)
31     }
32 }