/*

This Submission is to determine how many 120/240 min const. delivery point there are.

//info
120 req. steps <= 5

test_case 1,6 == 50 delivery point
		 2,3,4,5,7 == 80 delivery point
				8,9 == 90 delivery point
				10 > 100 delivery point
*/
#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <cstring>
#include <queue>
#include <stack>
#include <math.h>
#include <iterator>
#include <vector>
#include <string>
#include <set>
#include <math.h>
#include <iostream>
#include <random>
#include<map>
#include <iomanip>
#include <time.h>
#include <stdlib.h>
#include <list>
#include <typeinfo>
#include <list>
#include <set>
#include <cassert>
#include<fstream>
#include <unordered_map>
#include <cstdlib>
#include <complex>
#include <cctype>
#include <bitset>
using namespace std;
typedef string::const_iterator State;
#define Ma_PI 3.141592653589793
#define eps 0.00000001
#define LONG_INF 1e18
#define GOLD 1.61803398874989484820458
#define MAX_MOD 1000000007
#define MOD 1000000007
#define seg_size 262144
#define REP(a,b) for(long long a = 0;a < b;++a)

unsigned long xor128() {
	static unsigned long x = time(NULL), y = 362436069, z = 521288629, w = 88675123;
	unsigned long t = (x ^ (x << 11));
	x = y; y = z; z = w;
	return (w = (w ^ (w >> 19)) ^ (t ^ (t >> 8)));
}
long long func(long long n, long long a) {
	pair<long long, long long> now;
	now.second = min(n, a-1);
	now.first = max(1LL, a - n);
	return max(0LL,now.second - now.first +1LL);
}
long long go[300000];
long long base[300000];
int main() {
	long long n;
	cin >> n;
	REP(i, n) {
		long long a;
		cin >> a;
		base[i] = a;
		for (int q = 0; q < 24; ++q) {
			go[i + q] += a;
		}
	}
	long long now_ans = 0;
	for (int i = 0; i < 200050; ++i) {
		now_ans = max(now_ans, go[i]);
	}
	int query;
	cin >> query;
	vector<long long> ans;
	REP(i, query) {
		long long a, b;
		cin >> a >> b;
		a--;
		for (int q = 0; q < 24; ++q) {
			go[a + q] += b - base[a];
			now_ans = max(now_ans, go[a + q]);
		}
		ans.push_back(now_ans);
		base[a] = b;
	}
	REP(i, ans.size()) {
		cout << ans[i] << endl;
	}
}