Codeforces Round #367 (Div. 2) [B] Interesting drink

 Interesting drink

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.

Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of shops in the city that sell Vasiliy's favourite drink.

The second line contains n integers xi (1 ≤ xi ≤ 100 000) — prices of the bottles of the drink in the i-th shop.

The third line contains a single integer q (1 ≤ q ≤ 100 000) — the number of days Vasiliy plans to buy the drink.

Then follow q lines each containing one integer mi (1 ≤ mi ≤ 109) — the number of coins Vasiliy can spent on the i-th day.

Output

Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.

Example

input

53 10 8 6 114110311

output

0415

Note

On the first day, Vasiliy won't be able to buy a drink in any of the shops.

On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.

On the third day, Vasiliy can buy a drink only in the shop number 1.

Finally, on the last day Vasiliy can buy a drink in any shop.

题意:给你n个数，在给你m个数x1,x2.....xm，求出对于每一个xi，在n个数中小于等于xi的个数

这道题要用STL中的upper_bound,升序排列后upper_bound返回大于这个数的第一个数的下标，然后就非常好搞了

顺便总结一下lower_bound与upper_bound:

升序排列：

iterator lower_bound( const key_type &key ): 返回一个迭代器，指向键值>= key的第一个元素.

iterator upper_bound( const key_type &key ): 返回一个迭代器，指向键值 第一个>key的元素。

降序排列：

iterator lower_bound( const key_type &key ): 返回一个迭代器，指向键值<= key的第一个元素。

iterator upper_bound( const key_type &key ):返回一个迭代器，指向键值<key的最后一个元素的后一个元素

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cstdlib>#include<cmath>using namespace std;const int maxn=1e5+5,inf=1e9;inline void _read(int &x){    char t=getchar();bool sign=true;    while(t<'0'||t>'9')    {if(t=='-')sign=false;t=getchar();}    for(x=0;t>='0'&&t<='9';t=getchar())x=x*10+t-'0';    if(!sign)x=-x;}int n,m,s[maxn],num[maxn];struct wk{int x,id;}g[maxn];int main(){_read(n);int i,j=1;for(i=1;i<=n;i++)_read(s[i]);_read(m);for(i=1;i<=m;i++){_read(g[i].x);g[i].id=i;}sort(s+1,s+1+n);for(i=0;i+5<=maxn;i++){int pos=upper_bound(s+1,s+1+n,i)-s;num[i]=pos-1;}for(i=1;i<=m;i++)if(g[i].x>=s[n])printf("%d\n",n);else printf("%d\n",num[g[i].x]);}