Codeforces Round #428 (Div. 2)：D. Winter is here（组合数公式）

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

题意：

如果一个子序列的GCD为1，那么这个子序列的价值为0，否则子序列价值为子序列长度*子序列GCD

给出n个数，求这n个数所有子序列的价值和

题解：

考虑暴力每一个GCD，求出这个GCD对答案的贡献

如果枚举到x，n个数中有sum[x]个数是x的倍数，那么x对答案的贡献就是

但这可能会算重复，因为由全是x倍数组成的子序列它的GCD不一定是x啊

所以sum[x]还要减掉所有的sum[y]，其中y为x的倍数

那么令val[x]为x对答案的贡献，就有

所以枚举GCD时要倒过来枚举

最后求一波和就是答案

#include
#define mod 1000000007
#define LL long long
LL cnt[1000005], sum[1000005];
LL Pow(LL a, LL b)
{
	LL now;
	now = 1;
	while(b)
	{
		if(b%2)
			now = now*a%mod;
		a = a*a%mod;
		b /= 2;
	}
	return now;
}
int main(void)
{
	LL ans, i, j, n, x;
	scanf("%lld", &n);
	for(i=1;i<=n;i++)
	{
		scanf("%lld", &x);
		cnt[x]++;
	}
	ans = 0;
	for(i=1000000;i>=2;i--)
	{
		x = 0;
		for(j=i;j<=1000000;j+=i)
		{
			sum[i] -= sum[j];
			x += cnt[j];
		}
		sum[i] += x*Pow(2, x-1)%mod;
		ans = ((ans+sum[i]*i)%mod+mod)%mod;
	}
	printf("%lld\n", ans);
	return 0;
}