CF 839 D

D. Winter is here

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Winter is here at the North and the White Walkers are close. John Snow has an army consisting of n soldiers. While the rest of the world is fighting for the Iron Throne, he is going to get ready for the attack of the White Walkers.

He has created a method to know how strong his army is. Let the i-th soldier’s strength be ai. For some k he calls i1, i2, ..., ik a clan if i1 < i2 < i3 < ... < ik and gcd(ai1, ai2, ..., aik) > 1 . He calls the strength of that clan k·gcd(ai1, ai2, ..., aik). Then he defines the strength of his army by the sum of strengths of all possible clans.

Your task is to find the strength of his army. As the number may be very large, you have to print it modulo 1000000007 (109 + 7).

Greatest common divisor (gcd) of a sequence of integers is the maximum possible integer so that each element of the sequence is divisible by it.

Input

The first line contains integer n (1 ≤ n ≤ 200000) — the size of the army.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000000) — denoting the strengths of his soldiers.

Output

Print one integer — the strength of John Snow's army modulo 1000000007 (109 + 7).

Examples

input

3
3 3 1

output

12

input

4
2 3 4 6

output

39

Note

In the first sample the clans are {1}, {2}, {1, 2} so the answer will be 1·3 + 1·3 + 2·3 = 12

 

Solution:

首先有f(n)=Σi*C(i,n)(1<=i<=n)=i*2^i-1.

设ans[i]为gcd=i时的答案.

这要怎么求呢,首先什么是比较好求的呢?如果我们问的不是gcd,而是能整除(有哪些组都能被i整除的数,再分别乘以它们的元素个数),那不是很妙?

如果这样,我们不是只要求出有多少数能整除i,设其个数为n,那么答案就是f(n),不是吗?

其实就算问题是gcd,我们依然可以这么算,只不过会多计入gcd=2*i的,gcd=3*i的......那减去不就好了?反正是调和级数一个log而已是吧!

那么问题就结束了ans[i]=f[i]-Σans[j*i]

Code:

 

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
const int MAXN=200000;
const int MAXA=1000000;
const int MOD=1e9+7;
int n,a[MAXN+10];
int cnt[MAXA+10],f[MAXA+10],pow2[MAXN+10],ans[MAXA+10];
int AddMod(int a,int b){a+=b;return a>=MOD?a-MOD:a;}
int MinusMod(int a,int b){a-=b;return a<0?a+MOD:a;}
void init()
{
	pow2[0]=1;
	for(int i=1;i<=n;++i)++cnt[a[i]],pow2[i]=AddMod(pow2[i-1],pow2[i-1]);
	for(int i=1;i<=MAXA;++i)
	{
		for(int j=1;(ll)j*i<=MAXA;++j)
			f[i]+=cnt[j*i];
		if(f[i])f[i]=(ll)f[i]*pow2[f[i]-1]%MOD;
	}
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i)scanf("%d",a+i);
	init();
	int res=0;
	for(int i=MAXA;i>=2;i--)
	{
		ans[i]=f[i];
		for(int j=2;(ll)j*i<=MAXA;++j)
			ans[i]=MinusMod(ans[i],ans[j*i]);
		res=AddMod(res,(ll)i*ans[i]%MOD);
	}
	printf("%d",res);
	return 0;
}