Codeforces-839D Winter is here（容斥原理）

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 ifi1 < 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

题意：有一个序列A，设子序列的最小公约数为x=gcd(b1,b2,...,bk)，x>1时定义这个子序列的值为x*k，求A序列中所有子序列值的和

题解：容斥

枚举公约数x，找出x的倍数的个数n，那么x对答案的贡献为x*(1*C(n,1)+2*C(n,2)+...+n*C(n,n))

根据组合数公式i*C(n,i)=n*C(n-1,i-1)，上式可转化为x*n*2^(n-1)

不过这个贡献会有重复的，因此要减去gcd为x的倍数时的贡献，只要从大到小枚举x即可。

#include
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define x first
#define y second
using namespace std;
typedef long long LL;
typedef pair PII;
const int MX = 1e6 + 5;
const LL mod = 1e9 + 7;
LL dp[MX], t[MX];
int a[MX], cnt[MX];
int main() {
    int n, mx = 0;
    //freopen("in.txt", "r", stdin);
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        cnt[a[i]]++;
        mx = max(mx, a[i]);
    }
    t[0] = 1;
    for (int i = 1; i <= mx; i++) t[i] = t[i - 1] * 2 % mod;
    LL ans = 0;
    for (int k = mx; k > 1; k--) {
        int num = 0;
        for (int i = k; i <= mx; i += k) num += cnt[i];
        if (!num) continue;
        dp[k] = num * t[num - 1] % mod;
        for (int i = k * 2; i <= mx; i += k) dp[k] = (dp[k] - dp[i] + mod) % mod;
        ans = (ans + k * dp[k]) % mod;
    }
    printf("%I64d\n", ans);
    return 0;
}