CodeForces - 272C Dima and Staircase （线段树区间更新）

题意：

见以下样例，给出 5 个区间，每个区间的高度已知。一共 4 次操作。每次操作都是从最左边开始向下垒一个宽为 w 高为h 的木块，过程见下图。

问每次垒木块的高度是多少？

Input

Output

The first sample are shown on the picture.

思路：

线段树区间更新裸题，注意爆int

代码：

#include <bits/stdc++.h>using namespace std;
#define mi (l+r)/2;
#define ls l,mid,rt*2
#define rs mid+1,r,rt*2+1
const int MAXN=1e5;
long long tree[MAXN*4],lazy[MAXN*4];
long long n,m,st,en,ans,w,h;
void push_down(int rt){if(lazy[rt]){lazy[rt*2]=lazy[rt*2+1]=lazy[rt];tree[rt*2]=tree[rt*2+1]=lazy[rt];lazy[rt]=0;}return ;
}
void push_up(int rt){tree[rt]=max(tree[rt*2],tree[rt*2+1]);return ;
}
void build(int l,int r,int rt){if(l==r){scanf("%I64d",&tree[rt]);return ;}int mid=mi;build(ls);build(rs);push_up(rt);return ;
}
void update(int l,int r,int rt){if(st<=l&&r<=en){tree[rt]=lazy[rt]=ans;return ;}int mid=mi;push_down(rt);if(st<=mid) update(ls);if(mid<en) update(rs);push_up(rt);
}
void query(int l,int r,int rt){if(st<=l&&r<=en){ans=max(tree[rt],ans);return ;}int mid=mi;push_down(rt);if(st<=mid) query(ls);if(mid<en) query(rs);return ;
}
int main()
{while(scanf("%I64d",&n)!=-1){memset(tree,0,sizeof(tree));memset(lazy,0,sizeof(lazy));build(1,n,1);scanf("%I64d",&m);st=1;while(m--){scanf("%I64d%I64d",&w,&h);en=w;ans=0;query(1,n,1);printf("%I64d\n",ans);ans+=h;update(1,n,1);}}
}

Dima's got a staircase that consists of n stairs. The first stair is at height a₁, the second one is at a₂, the last one is at a_n (1 ≤ a₁ ≤ a₂ ≤ ... ≤ a_n).

Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width w_i and height h_i. Dima throws each box vertically down on the first w_i stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., w_i. Each thrown box flies vertically down until at least one of the two following events happen:

the bottom of the box touches the top of a stair;
the bottom of the box touches the top of a box, thrown earlier.

We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width w_i cannot touch the stair number w_i + 1.

You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.

Input

The first line contains integer n (1 ≤ n ≤ 10⁵) — the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of nintegers, a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹; a_i ≤ a_i + 1).

The next line contains integer m (1 ≤ m ≤ 10⁵) — the number of boxes. Each of the following m lines contains a pair of integers w_i, h_i (1 ≤ w_i ≤ n; 1 ≤ h_i ≤ 10⁹) — the size of the i-th thrown box.

The numbers in the lines are separated by spaces.

Output

Print m integers — for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

Example

Input

Output

Input

Output

1
3

Input

Output

Note

The first sample are shown on the picture.

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