# GeetCode Hub

We have an array `nums` of integers, and an array `queries` of queries.

For the `i`-th query `val = queries[i], index = queries[i]`, we add val to `nums[index]`.  Then, the answer to the `i`-th query is the sum of the even values of `A`.

(Here, the given `index = queries[i]` is a 0-based index, and each query permanently modifies the array `nums`.)

Return the answer to all queries.  Your `answer` array should have `answer[i]` as the answer to the `i`-th query.

Example 1:

```Input: nums = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
Output: [8,6,2,4]
Explanation:
At the beginning, the array is [1,2,3,4].
After adding 1 to nums, the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to nums, the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to nums, the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to nums, the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.
```

Note:

1. `1 <= nums.length <= 10000`
2. `-10000 <= nums[i] <= 10000`
3. `1 <= queries.length <= 10000`
4. `-10000 <= queries[i] <= 10000`
5. `0 <= queries[i] < nums.length`

class Solution { public int[] sumEvenAfterQueries(int[] nums, int[][] queries) { } }