<p>You are given two <strong>positive</strong> integers <code>startPos</code> and <code>endPos</code>. Initially, you are standing at position <code>startPos</code> on an <strong>infinite</strong> number line. With one step, you can move either one position to the left, or one position to the right.</p>

<p>Given a positive integer <code>k</code>, return <em>the number of <strong>different</strong> ways to reach the position </em><code>endPos</code><em> starting from </em><code>startPos</code><em>, such that you perform <strong>exactly</strong> </em><code>k</code><em> steps</em>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>Two ways are considered different if the order of the steps made is not exactly the same.</p>

<p><strong>Note</strong> that the number line includes negative integers.</p>

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

<pre>
<strong>Input:</strong> startPos = 1, endPos = 2, k = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can reach position 2 from 1 in exactly 3 steps in three ways:
- 1 -&gt; 2 -&gt; 3 -&gt; 2.
- 1 -&gt; 2 -&gt; 1 -&gt; 2.
- 1 -&gt; 0 -&gt; 1 -&gt; 2.
It can be proven that no other way is possible, so we return 3.</pre>

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

<pre>
<strong>Input:</strong> startPos = 2, endPos = 5, k = 10
<strong>Output:</strong> 0
<strong>Explanation:</strong> It is impossible to reach position 5 from position 2 in exactly 10 steps.
</pre>

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

<ul>
	<li><code>1 &lt;= startPos, endPos, k &lt;= 1000</code></li>
</ul>

def retrieve(d: int, k: int, memo: List[List[int]]):
    if k < d:
        return 0
    elif k == d:
        return 1
    return memo[d][k]
    
def solve_from_bottom(d: int, k: int, memo: List[List[int]]) -> int:
    return (retrieve(abs(d-1),k-1,memo) + retrieve(abs(d+1),k-1,memo)) % (10**9 + 7)

class Solution:
    def numberOfWays(self, startPos: int, endPos: int, k: int) -> int:
        d = abs(endPos - startPos)

        memo = [[-1 for i in range(k+2)] for j in range(k+2)]
        
        if k < d:
            return 0
        elif k == d:
            return 1
        elif k % 2 != d % 2:
            return 0
        
        for c in range(1,k // 2 + 1):
            for i in range(c):
                t = solve_from_bottom(0+2*i,2*c,memo)
                memo[0+2*i][2*c] = t
            for i in range(c):
                t = solve_from_bottom(1+2*i,1+2*c,memo)
                memo[1+2*i][1+2*c] = t
            
            if memo[d][k] != -1:
                return memo[d][k]

            
            

561