Codechef ForbiddenSum

Mike likes to invent new functions. The latest one he has invented is called ForbiddenSum. Let's consider how it can be calculated:

You are given a set S of positive integers. The integers are not necessary distinct. ForbiddenSum of S equals to the minimal non-negative integer, that can't be returned by the algorithm described below:

 

• Choose a random subset S' of S(it can be empty as well);
• Calculate the sum of all the numbers of S' and assign it to the variable T;
• Return the value of the variable T.

 

I.e. if S = {1, 1, 3, 7}, the algorithm can return 0(S' = {}), 1(S' = {1}), 2(S' = {1, 1}), 3(S' = {3}), 4(S' = {1, 3}), 5(S' = {1, 1, 3}), but it can't return 6. So, ForbiddenSum of S equals to 6.

Inventing of new function is not the only Mike's hobby. Also, he likes to collect rare and unusual arrays. He is going to share with you one of them.

Formally, Mike gives you one array A, consisting of N positive integers. After that, he asks you M questions, two integers L_i and R_i describe i'th question: What does ForbiddenSum of S={A_Li, A_Li+1, ..., A_Ri-1, A_Ri} equal to?

Input

The first line contains the only integer N. The second line contains N integers - the array A. A is 1-indexed.

The third line contains the only integer M. The following M lines contain two integer numbers 1 ≤ L_i ≤ R_i ≤ N each.

 

Output

Output should consists of M lines. The i'th line should contains the answer to the i'th question.

 

Constraints

1 ≤ N, M ≤ 100,000

1 ≤ A_i ≤ 10⁹

1 ≤ A₁ + A₂ + ... + A_N ≤ 10⁹

 

Example

Input:
5
1 2 4 9 10
5
1 1
1 2
1 3
1 4
1 5

Output:
2
4
8
8
8

 

Explanation

In the example test there are M=5 questions. We won't describe all of them, only two ones.

The first question

In the first test case S equals to {1}. The answer is 2, because we can recieve 1 in case the algorithm chooses S' = {1}. But there are no chances to receive 2.

The second question

In the first test case S equals to {1, 2}. The answer is 4, because we can recieve 1 in case the algorithm chooses S' = {1}, 2 in case the algorithm chooses S' = {2} and 3 in case the algorithm chooses S' = {1, 2}. But there are no chances to receive 4.

 

中文题面在这： Mandarin Chinese 

 

有一个性质是我随便口胡的虽然不知道怎么证明但是却用它A了这个题。

那就是如果集合按数字大小排序之后，i是最小的满足a[1]+a[2]+...a[i]<a[i+1]-1,那么这个集合的forbidden sum为a[i+1]-1.

这就相当于表示不了a[i+1]-1这个数了。

而如果>=的话那么是可以接上的。

 

 

于是就可以暴力做啦，对于每次询问，设最大能表示的数为num。

一开始num都为0，然后每次查找区间内权值∈[1,num+1]的数的和，然后把num设为这个值。

当然如果这个值==num的话，说明无法扩展了，输出num+1然后去管下一个询问就好了。

 

这个的话用主席树比较好(当然你也可以用其他的)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ll long long
#define maxn 100005
using namespace std;
const int inf=1<<30;
struct node{
    int lc,rc;
    int s;
}nil[maxn*37];
int n,m,ky,a[maxn],cnt=0;
int rot[maxn];
int le,ri,S,T,ans,pre;

int update(int u,int l,int r){
    int ret=++cnt;
    nil[ret]=nil[u];
    nil[ret].s+=le;
    if(l==r) return ret;
    
    int mid=l+r>>1;
    if(le<=mid) nil[ret].lc=update(nil[ret].lc,l,mid);
    else nil[ret].rc=update(nil[ret].rc,mid+1,r);
    
    return ret;
}

int query(int x,int y,int l,int r){
    if(l>=le&&r<=ri) return nil[y].s-nil[x].s;
    int mid=l+r>>1,an=0;
    if(le<=mid) an+=query(nil[x].lc,nil[y].lc,l,mid);
    if(ri>mid) an+=query(nil[x].rc,nil[y].rc,mid+1,r);
    return an; 
}

inline void prework(){
    rot[0]=nil->lc=nil->rc=0;
    nil->s=0;
    
    for(int i=1;i<=n;i++){
        le=a[i];
        rot[i]=update(rot[i-1],1,1e9);
    }
}

int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",a+i);
    
    prework();
    
    scanf("%d",&m);
    while(m--){
        scanf("%d%d",&S,&T);
        pre=-1,ans=0;
        while(ans>pre){
            pre=ans;
            le=1,ri=ans+1;
            ans=query(rot[S-1],rot[T],1,1e9);
        }
        
        printf("%d\n",ans+1);
    }
    
    return 0;
}