We have an array `nums`

of integers, and an array `queries`

of queries.

For the `i`

-th query `val = queries[i][0], index = queries[i][1]`

, 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][1] 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[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8. After adding -3 to nums[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6. After adding -4 to nums[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2. After adding 2 to nums[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.

**Note:**

`1 <= nums.length <= 10000`

`-10000 <= nums[i] <= 10000`

`1 <= queries.length <= 10000`

`-10000 <= queries[i][0] <= 10000`

`0 <= queries[i][1] < nums.length`

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