Codeforces 706B Interesting drink-白红宇

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 x_i 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 m_i 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 x_i (1 ≤ x_i ≤ 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 m_i (1 ≤ m_i ≤ 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

5 3 10 8 6 11 4 1 10 3 11

Output

0 4 1 5

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.

题目链接：

分析：二分的简单应用吧！在区间内查找上限下限，然后根据这些求出即可！

下面给出AC代码：

1 #include 
       
         2 using namespace std; 3 int main() 4 { 5     int a[100010]; 6     int n,i,m,num; 7     while(scanf("%d",&n)!=EOF) 8     { 9         for(i=1;i<=n;i++)10             scanf("%d",&a[i]);11         sort(a+1,a+1+n);12         scanf("%d",&m);13         while(m--)14         {15             scanf("%d",&num);16             if(num
        
         =a[n])19                 printf("%d\n",n);20             else21             {22                 int left=1,right=n;23                 int mid,ans;24                 while(left<=right)25                 {26                     mid=(left+right)/2;27                     if(a[mid]<=num)28                     {29                         ans=mid;30                         left=mid+1;31                     }32                     else right=mid-1;33                 }34                 printf("%d\n",ans);35             }36         }37     }38     return 0;39 }