<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>

JUMPS = {
    1: [6, 8],
    2: [7, 9],
    3: [4, 8],
    4: [0, 3, 9],
    5: [],
    6: [0, 1, 7],
    7: [2, 6],
    8: [1, 3],
    9: [2, 4],
    0: [4, 6]
}
M = 10 ** 9 + 7

class Solution:
    def knightDialer(self, n: int) -> int:
        # Maps from starting digit -> length of string -> # of possible phone numbers
        moves = {}
        for x in range(10):
            moves[x] = {1: 1}
        
        for k in range(2, n + 1):
            for x in range(10):
                possible_phones = 0
                # Iterate over all possible starting positions (this is the first number)
                for j in JUMPS[x]:
                    # Jump from x -> j
                    possible_phones += moves[j][k - 1] % M
                
                moves[x][k] = possible_phones % M
        
        answer = 0
        # print(moves)
        """
        04
        06

        494
        492
        434
        438
        404
        406

        """
        for x in range(10):
            answer += moves[x][n]
            answer = answer % M
        return answer
20344