<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 knightDialer(int n) {
        vector<vector<vector<int> > > dp(n+1, vector<vector<int> > (4, vector<int> (3,0)));
        int mod = 1000000007;
        for(int i=0;i<4;i++){
            for(int j=0;j<3;j++){
                dp[0][i][j]=1;
            }
        }
        for(int i=0;i<n;i++){
            dp[i][3][0]=-1;
            dp[i][3][2]=-1;
        }
        vector<vector<int> > dir = {{2,1},{-2,1},{2,-1},{-2,-1},{1,-2},{1,2},{-1,2},{-1,-2}};
        for(int i=1;i<n;i++){
            for(int j=0;j<4;j++){
                for(int k=0;k<3;k++){
                    if(dp[i][j][k]==-1) continue;
                    for(int l=0;l<8;l++){
                        int x = j+dir[l][0];
                        int y = k+dir[l][1];
                        if(x>=0 && y>=0 && x<4 && y<3 && dp[i-1][x][y]!=-1){
                            dp[i][j][k]= (dp[i][j][k]%mod + dp[i-1][x][y]%mod)%mod;
                        }
                    }
                }
            }
        }
        int ans =0;
        for(int i=0;i<4;i++){
            for(int j=0;j<3;j++){
                if(dp[n-1][i][j]!=-1){
                    ans= (ans+dp[n-1][i][j])%mod;
                }
            }
        }
        return ans;
    }
};
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