<p>You are given an array <code>nums</code> consisting of <strong>positive</strong> integers and an integer <code>k</code>.</p>

<p><strong>Partition</strong> the array into two ordered <strong>groups</strong> such that each element is in exactly <strong>one</strong> group. A partition is called great if the <strong>sum</strong> of elements of each group is greater than or equal to <code>k</code>.</p>

<p>Return <em>the number of <strong>distinct</strong> great partitions</em>. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>Two partitions are considered distinct if some element <code>nums[i]</code> is in different groups in the two partitions.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4], k = 4
<strong>Output:</strong> 6
<strong>Explanation:</strong> The great partitions are: ([1,2,3], [4]), ([1,3], [2,4]), ([1,4], [2,3]), ([2,3], [1,4]), ([2,4], [1,3]) and ([4], [1,2,3]).
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,3,3], k = 4
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no great partitions for this array.
</pre>

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

<pre>
<strong>Input:</strong> nums = [6,6], k = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong> We can either put nums[0] in the first partition or in the second partition.
The great partitions will be ([6], [6]) and ([6], [6]).
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length, k &lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def countPartitions(self, nums: List[int], k: int) -> int:
        MOD = 10**9 + 7
        n = len(nums)
        total_sum = sum(nums)
        if total_sum < 2 * k:
            return 0
        
        dp = [[0] * (k + 1) for _ in range(n + 1)]
        dp[0][0] = 1
        
        for i in range(1, n + 1):
            for j in range(k + 1):
                dp[i][j] = dp[i - 1][j]
                if j >= nums[i - 1]:
                    dp[i][j] += dp[i - 1][j - nums[i - 1]]
                dp[i][j] %= MOD
        
        result = 0
        for j in range(k):
            result += dp[n][j]
            result %= MOD
        
        return (pow(2, n, MOD) - 2 * result) % MOD
562