<p>You have <code>n</code> dice, and each dice has <code>k</code> faces numbered from <code>1</code> to <code>k</code>.</p>

<p>Given three integers <code>n</code>, <code>k</code>, and <code>target</code>, return <em>the number of possible ways (out of the </em><code>k<sup>n</sup></code><em> total ways) </em><em>to roll the dice, so the sum of the face-up numbers equals </em><code>target</code>. Since the answer may be too large, return it <strong>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, k = 6, target = 3
<strong>Output:</strong> 1
<strong>Explanation:</strong> You throw one die with 6 faces.
There is only one way to get a sum of 3.
</pre>

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

<pre>
<strong>Input:</strong> n = 2, k = 6, target = 7
<strong>Output:</strong> 6
<strong>Explanation:</strong> You throw two dice, each with 6 faces.
There are 6 ways to get a sum of 7: 1+6, 2+5, 3+4, 4+3, 5+2, 6+1.
</pre>

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

<pre>
<strong>Input:</strong> n = 30, k = 30, target = 500
<strong>Output:</strong> 222616187
<strong>Explanation:</strong> The answer must be returned modulo 10<sup>9</sup> + 7.
</pre>

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

<ul>
	<li><code>1 &lt;= n, k &lt;= 30</code></li>
	<li><code>1 &lt;= target &lt;= 1000</code></li>
</ul>

class Solution:
    def numRollsToTarget(self, d: int, f: int, target: int) -> int:
        MOD = 10**9 + 7
        
        # dp[i][j] means the number of ways to roll i dice to sum up to j
        dp = [[0] * (target + 1) for _ in range(d + 1)]
        
        # Base case: 1 way to achieve target 0 with 0 dice (do nothing)
        dp[0][0] = 1
        
        # Iterate over the number of dice (from 1 to d)
        for i in range(1, d + 1):
            # Iterate over all possible targets (from 1 to target)
            for j in range(1, target + 1):
                # Iterate over each face of the dice (from 1 to f)
                for k in range(1, f + 1):
                    if j - k >= 0:  # Only valid if target is achievable
                        dp[i][j] = (dp[i][j] + dp[i - 1][j - k])
        
        return dp[d][target]% MOD
17400