<p>The chess knight has a <strong>unique movement</strong>,&nbsp;it may move two squares vertically and one square horizontally, or two squares horizontally and one square vertically (with both forming the shape of an <strong>L</strong>). The possible movements of chess knight are shown in this diagram:</p>

<p>A chess knight can move as indicated in the chess diagram below:</p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/18/chess.jpg" style="width: 402px; height: 402px;" />
<p>We have a chess knight and a phone pad as shown below, the knight <strong>can only stand on a numeric cell</strong>&nbsp;(i.e. blue cell).</p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/18/phone.jpg" style="width: 242px; height: 322px;" />
<p>Given an integer <code>n</code>, return how many distinct phone numbers of length <code>n</code> we can dial.</p>

<p>You are allowed to place the knight <strong>on any numeric cell</strong> initially and then you should perform <code>n - 1</code> jumps to dial a number of length <code>n</code>. All jumps should be <strong>valid</strong> knight jumps.</p>

<p>As the answer may be very large, <strong>return the answer modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 10
<strong>Explanation:</strong> We need to dial a number of length 1, so placing the knight over any numeric cell of the 10 cells is sufficient.
</pre>

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

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 20
<strong>Explanation:</strong> All the valid number we can dial are [04, 06, 16, 18, 27, 29, 34, 38, 40, 43, 49, 60, 61, 67, 72, 76, 81, 83, 92, 94]
</pre>

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

<pre>
<strong>Input:</strong> n = 3131
<strong>Output:</strong> 136006598
<strong>Explanation:</strong> Please take care of the mod.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 5000</code></li>
</ul>

class Solution {
public:
    int mod = 1e9 + 7;
    int arr[8] = {1, 1, 2, 2, -2, -2, -1, -1};
    int brr[8] = {2, -2, 1, -1, 1, -1, 2, -2};
    int rec(int i, int j, vector<vector<int>>& mat, vector<vector<vector<int>>>& dp, int n){
        if(n <= 0 || (i == 3 && j == 0) || (i == 3 && j == 2) || i < 0 || j < 0 || i >= 4 || j >= 3) return 0;
        if(n == 1) return 1;
        if(dp[i][j][n] != -1) return dp[i][j][n];
        long long temp = 0;
        for(int a = 0; a < 8; a++){
            int add = rec(i + arr[a], j + brr[a], mat, dp, n - 1);
            temp = (temp + add) % mod;
        }
        int ans = temp % mod;
        return dp[i][j][n] = ans;
    }
    int knightDialer(int n) {
        if(n == 1) return 10;
        vector<vector<int>> mat = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}, {-1, 0, -1}};
        vector<vector<vector<int>>> dp(4, vector<vector<int>>(3, vector<int>(5001, -1)));
        long long temp = 0;
        for(int i = 0; i < 4; i++){
            for(int j = 0; j < 3; j++) temp += rec(i, j, mat, dp, n);
            temp %= mod;
        }
        int ans = temp % mod;
        return ans;
    }
};
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