Codeforces475D-CGCDSSQ

CGCDSSQ

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r)such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.

 is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.

Input

The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).

The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).

Output

For each query print the result in a separate line.

Examples

input

32 6 3512346

output

12201

input

710 20 3 15 1000 60 16101234561020601000

output

14022202211

题意：给你有n个数的序列，有m次询问，求出有多少个区间的gcd为这个值

解题思路：可以先预处理出所有区间gcd的值，对于每次询问直接输出

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n, m, x;map<int, LL>mp;int gcd(int x, int y){return x ? gcd(y%x, x) : y;}int main(){while (~scanf("%d", &n)){mp.clear();stack<pair<int, LL> >s1, s2;for (int i = 1; i <= n; i++){scanf("%d", &x);s1.push(make_pair(x, 1));while (!s1.empty()) s2.push(make_pair(gcd(s1.top().first, x), s1.top().second)), s1.pop();while (!s2.empty()){pair<int, LL>pre = s2.top(); s2.pop();mp[pre.first] += pre.second;if (!s1.empty() && pre.first == s1.top().first) pre.second += s1.top().second, s1.pop();s1.push(pre);}}scanf("%d", &m);while (m--){scanf("%d", &x);printf("%lld\n", mp[x]);}}return 0;}