You are given a sequence A[1], A[2], ..., A[N] . ( |A[i]| ≤ 15007 , 1 ≤ N ≤ 50000 ). A query is defined as follows:
Query(x,y) = Max { a[i]+a[i+1]+...+a[j] ; x ≤ i ≤ j ≤ y }.
Given M queries, your program must output the results of these queries.

Input
The first line of the input file contains the integer N.
In the second line, N numbers follow.
The third line contains the integer M.
M lines follow, where line i contains 2 numbers xi and yi.
Output
Your program should output the results of the M queries, one query per line.

Sample Input
3
-1 2 3
1
1 2
Sample Output
2

题意：

裸题，没有更新，只有查询。

 思路：

区间中维护一下值：

从左端点开始的连续的最大值lm

从右端点开始的连续最大值rm

区间的和sum。

区间中的连续最大值num。

那么更新操作为：

lm=max（左儿子的lm，左儿子的sum+右儿子的lm）

rm=max（右儿子的rm，右儿子的sum+左儿子的rm）

num=max（左儿子的num，右儿子的num，左儿子的rm+右儿子的lm）

询问操作也返回区间，对区间进行合并处理操作。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 50000+7;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
struct node
{
    int l;
    int r;
    ll num;
    ll lm;
    ll sum;
    ll rm;
}segment_tree[maxn<<2];
int n;
void pushup(int rt)
{
    segment_tree[rt].sum=segment_tree[rt<<1].sum+segment_tree[rt<<1|1].sum;
    segment_tree[rt].lm=max(segment_tree[rt<<1].lm,segment_tree[rt<<1].sum+segment_tree[rt<<1|1].lm);
    segment_tree[rt].rm=max(segment_tree[rt<<1|1].rm,segment_tree[rt<<1|1].sum+segment_tree[rt<<1].rm);
    segment_tree[rt].num=max(segment_tree[rt<<1].num,segment_tree[rt<<1|1].num);
    segment_tree[rt].num=max(segment_tree[rt].num,segment_tree[rt<<1].rm+segment_tree[rt<<1|1].lm);
}

void build(int rt,int l,int r)
{
    segment_tree[rt].l=l;
    segment_tree[rt].r=r;
    if(l==r)
    {
        scanf("%lld",&segment_tree[rt].num);
        segment_tree[rt].lm=segment_tree[rt].rm=segment_tree[rt].num;
        segment_tree[rt].sum=segment_tree[rt].num;
        return ;
    }
    int mid=(l+r)>>1;
    build(rt<<1,l,mid);
    build(rt<<1|1,mid+1,r);
    pushup(rt);
}

node ask(int rt,int l,int r)
{
    if(segment_tree[rt].l==l&&segment_tree[rt].r==r)
    {
        return segment_tree[rt];
    }
    int mid=(segment_tree[rt].r+segment_tree[rt].l)>>1;
    if(l>mid)
    {
        return ask(rt<<1|1,l,r);
    }else if(r<=mid)
    {
        return ask(rt<<1,l,r);
    }else
    {
        node res1=ask(rt<<1,l,mid);
        node res2=ask(rt<<1|1,mid+1,r);
        node res;
        res.sum=res1.sum+res2.sum;
        res.lm=max(res1.lm,res1.sum+res2.lm);
        res.rm=max(res2.rm,res2.sum+res1.rm);
        res.num=max(res1.num,res2.num);
        res.num=max(res.num,res1.rm+res2.lm);
        return res;
    }
}

int main()
{
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);
    scanf("%d",&n);
    build(1,1,n);
    int m;
    scanf("%d",&m);
    while(m--)
    {
        int x,y;
        scanf("%d %d",&x,&y);
        printf("%lld\n",ask(1,x,y).num);
    }
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}