# -*- coding: utf-8 -*-
"""
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.
"""

class Solution(object):
    def sumEvenAfterQueries(self, A, queries):
        """
        :type A: List[int]
        :type queries: List[List[int]]
        :rtype: List[int]
        """
        
        """ Time Limit Exceeded
        ans = []
        ans_append = ans.append
        for i in range(len(A)):
            val = queries[i][0]
            index  = queries[i][1]
            A[index] += val
            ans_append(sum(j for j in A if j % 2 == 0))
            
        return ans
        """
        
        S = sum(x for x in A if x % 2 == 0)
        ans = []
        
        for x, k in queries:
            if A[k] % 2 == 0:
                S -= A[k]
            A[k] += x
            if A[k] % 2 == 0:
                S += A[k]
            ans.append(S)
        
        return ans        
        
if __name__ == "__main__":
    
    A = [1, 2, 3, 4]
    solution = Solution()
    print(solution.sumEvenAfterQueries(A))