You are given an integer array nums of length n and a positive integer k.
The power of an array of integers is defined as the number of subsequences with their sum equal to k.
Return the sum of power of all subsequences of nums.
Since the answer may be very large, return it modulo 109 + 7.
\n
Example 1:
\n\nInput: nums = [1,2,3], k = 3
\n\nOutput: 6
\n\nExplanation:
\n\nThere are 5 subsequences of nums with non-zero power:
[1,2,3] has 2 subsequences with sum == 3: [1,2,3] and [1,2,3].[1,2,3] has 1 subsequence with sum == 3: [1,2,3].[1,2,3] has 1 subsequence with sum == 3: [1,2,3].[1,2,3] has 1 subsequence with sum == 3: [1,2,3].[1,2,3] has 1 subsequence with sum == 3: [1,2,3].Hence the answer is 2 + 1 + 1 + 1 + 1 = 6.
Example 2:
\n\nInput: nums = [2,3,3], k = 5
\n\nOutput: 4
\n\nExplanation:
\n\nThere are 3 subsequences of nums with non-zero power:
[2,3,3] has 2 subsequences with sum == 5: [2,3,3] and [2,3,3].[2,3,3] has 1 subsequence with sum == 5: [2,3,3].[2,3,3] has 1 subsequence with sum == 5: [2,3,3].Hence the answer is 2 + 1 + 1 = 4.
Example 3:
\n\nInput: nums = [1,2,3], k = 7
\n\nOutput: 0
\n\nExplanation: There exists no subsequence with sum 7. Hence all subsequences of nums have power = 0.
\n
Constraints:
\n\n1 <= n <= 1001 <= nums[i] <= 1041 <= k <= 100给你一个长度为 n 的整数数组 nums 和一个 正 整数 k 。
一个整数数组的 能量 定义为和 等于 k 的子序列的数目。
请你返回 nums 中所有子序列的 能量和 。
由于答案可能很大，请你将它对 109 + 7 取余 后返回。
\n\n
示例 1：
\n\n输入： nums = [1,2,3], k = 3
\n\n输出： 6
\n\n解释：
\n\n总共有 5 个能量不为 0 的子序列：
[1,2,3] 有 2 个和为 3 的子序列：[1,2,3] 和 [1,2,3] 。[1,2,3] 有 1 个和为 3 的子序列：[1,2,3] 。[1,2,3] 有 1 个和为 3 的子序列：[1,2,3] 。[1,2,3] 有 1 个和为 3 的子序列：[1,2,3] 。[1,2,3] 有 1 个和为 3 的子序列：[1,2,3] 。所以答案为 2 + 1 + 1 + 1 + 1 = 6 。
示例 2：
\n\n输入： nums = [2,3,3], k = 5
\n\n输出： 4
\n\n解释：
\n\n总共有 3 个能量不为 0 的子序列：
[2,3,3] 有 2 个子序列和为 5 ：[2,3,3] 和 [2,3,3] 。[2,3,3] 有 1 个子序列和为 5 ：[2,3,3] 。[2,3,3] 有 1 个子序列和为 5 ：[2,3,3] 。所以答案为 2 + 1 + 1 = 4 。
示例 3：
\n\n输入： nums = [1,2,3], k = 7
\n\n输出： 0
\n\n解释：不存在和为 7 的子序列，所以 nums 的能量和为 0 。
\n\n
提示：
\n\n1 <= n <= 1001 <= nums[i] <= 1041 <= k <= 100j with the sum of elements k, it contributes 2n - j to the answer.",
"Let dp[i][j] represent the number of subsequences in the subarray nums[0..i] which have a sum of j.",
"We can find the dp[i][k] for all 0 <= i <= n-1 and multiply them with 2n - j to get final answer."
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