We have an array A 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 A[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 A.) Return the answer to all queries. Your answer array should have answer[i] as the answer to the i-th query. Example 1: Input: A = [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 A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8. After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6. After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2. After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4. Note: 1 <= A.length <= 10000 -10000 <= A[i] <= 10000 1 <= queries.length <= 10000 -10000 <= queries[i][0] <= 10000 0 <= queries[i][1] < A.length