At the entrance to the university, there is a huge rectangular billboard of size h*w (h is its height and w is its width). The board is the place where all possible announcements are posted: nearest programming competitions, changes in the dining room menu, and other important information.

On September 1, the billboard was empty. One by one, the announcements started being put on the billboard.

Each announcement is a stripe of paper of unit height. More specifically, the i-th announcement is a rectangle of size 1 * wi.

When someone puts a new announcement on the billboard, she would always choose the topmost possible position for the announcement. Among all possible topmost positions she would always choose the leftmost one.

If there is no valid location for a new announcement, it is not put on the billboard (that’s why some programming contests have no participants from this university).

Given the sizes of the billboard and the announcements, your task is to find the numbers of rows in which the announcements are placed.
Input
There are multiple cases (no more than 40 cases).

The first line of the input file contains three integer numbers, h, w, and n (1 <= h,w <= 10^9; 1 <= n <= 200,000) - the dimensions of the billboard and the number of announcements.

Each of the next n lines contains an integer number wi (1 <= wi <= 10^9) - the width of i-th announcement.
Output
For each announcement (in the order they are given in the input file) output one number - the number of the row in which this announcement is placed. Rows are numbered from 1 to h, starting with the top row. If an announcement can’t be put on the billboard, output “-1” for this announcement.
Sample Input
3 5 5
2
4
3
3
3

题意：给出高为h，宽为w的海报墙，给出n张海报的宽度（高度都为1），每次放在满足条件的最左上角（不能溢出），问能放在第几行，不能放的话输出-1。

思路：一开始不知道怎么建树，因为h最大能达到1e9，，后来看了看别人博客，发现，，每张海报最坏的情况就是它自己占一行，所以最多也就n行，n范围是2e5，完全ok的。那么每个叶子节点代表一行，初始化为w，父节点保存子节点的最大剩余宽度就可以了。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int N = 2e5+10;
int sum[N<<2];
int h,w;
void pushup(int x)
{
	sum[x]=max(sum[x<<1],sum[x<<1|1]);
}
void build(int x,int l,int r)
{
	if(l==r)
	{
		sum[x]=w;return;
	}
	int mid=l+r>>1;
	build(x<<1,l,mid);
	build(x<<1|1,mid+1,r);
	pushup(x);
}
void update(int x,int l,int r,int tem,int val)
{
	if(l==r)
	{
		sum[x]-=val;return;
	}
	int mid=l+r>>1;
	if(tem<=mid)
		update(x<<1,l,mid,tem,val);
	else
		update(x<<1|1,mid+1,r,tem,val);
	pushup(x);
}
int query(int x,int l,int r,int height)   //其实可以把查询和更新写一块，但是我为了不出错，就分开写了
{
	if(l==r)
	{
		return l;   //每次返回小的
	}
	int mid=l+r>>1,ans=0;
	if(sum[x<<1]>=height)
		ans=query(x<<1,l,mid,height);
	else if(sum[x<<1|1]>=height)
		ans=query(x<<1|1,mid+1,r,height);
	return ans;
}
int main()
{
	int t,n;
	while(cin>>h>>w>>n)
	{
		 
		 if(h>n)   //高度大于n的话，多出来的部分就用不上了
		 	h=n; 
		 build(1,1,h);
		 while(n--)
		 {
		 	int height;
		 	scanf("%d",&height);		 	
			if(sum[1]<height)  //如果所有行的剩余高度中最大的都小于它，那么无法贴它
		 	{
		 		printf("-1\n");continue;
			}
			int x=query(1,1,h,height);
		 	printf("%d\n",x);update(1,1,h,x,height);;
		 }
	}
	return 0;
}