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".
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.
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.
5 3 10 8 6 11 4 1 10 3 11
0 4 1 5
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.
#include<stdio.h> #include<iostream> #include<string.h> #include<string> #include<ctype.h> #include<math.h> #include<set> #include<map> #include<vector> #include<queue> #include<bitset> #include<algorithm> #include<time.h> using namespace std; void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); } #define MS(x,y) memset(x,y,sizeof(x)) #define MC(x,y) memcpy(x,y,sizeof(x)) #define MP(x,y) make_pair(x,y) #define ls o<<1 #define rs o<<1|1 typedef long long LL; typedef unsigned long long UL; typedef unsigned int UI; template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; } template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; } const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f; int n; int a[N]; int q; int main() { while (~scanf("%d", &n)) { for (int i = 1; i <= n; ++i)scanf("%d", &a[i]); sort(a + 1, a + n + 1); scanf("%d", &q); while (q--) { int x; scanf("%d", &x); printf("%d\n", upper_bound(a + 1, a + n + 1, x) - a - 1); } } return 0; } /* 【trick&&吐槽】 upper_bound()是返回第一个>x的数的下标 lower_bound()是返回第一个>=x的数的下标 */