<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 xx[8] = {1,2,-1,-2,1,-2,-1,2};
    int yy[8] = {2,1,-2,-1,-2,1,2,-1};

    int knightDialer(int n) {
        vector<vector<long long>> dp(n+1, vector<long long>(12, 0));

        long long ans = 0;

        long long MOD = 1e9+7;

        for(int i=1;i<=n;i++){
            for(int j=0;j<12;j++){
                if(j == 9 || j == 11){
                    dp[i][j] = 0;
                    continue;
                }

                if(i == 1){
                    dp[i][j] = 1;
                }
                else{
                    int y = j%3;
                    int x = (j-y)/3;

                    for(int k=0;k<8;k++){
                        int nx = x + xx[k];
                        int ny = y + yy[k];

                        if(nx <0 || ny<0 || nx>=4 || ny>=3){
                            continue;
                        }

                        dp[i][j] += dp[i-1][3*nx+ny];

                        dp[i][j]%=MOD;
                    }
                }

                
                if(i == n){
                    ans += dp[i][j];
                    ans%=MOD;
                }
            }
        }

        return ans;
    }
};
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