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.

class Solution {
public:
vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {
int initval = ;
vector<int> ret;
for(int v : A) {
if(v % == ) initval += v;
}
for(auto v : queries) {
int before = A[v[]];
A[v[]] += v[];
if(before % == ) initval -= before;
if(A[v[]] % == ) initval += A[v[]];
ret.push_back(initval);
}
return ret;
}
};

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