Codeforces-822D Winter is here（DP）

最新推荐文章于 2021-02-28 14:49:16 发布

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本文介绍了一个算法问题，模拟《权力的游戏》中约翰·雪诺评估其军队实力的方式。通过定义特定的士兵族群及其强度，该算法需计算所有可能族群的总强度，并输出一个巨大的数值模10^9 + 7的结果。

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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

#include<bits/stdc++.h>
#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<int, int> PII;
const int MX = 5e6 + 5;
const LL mod = 1e9 + 7;
int a[MX], arr[MX];
void init() {
    for (int i = 2; i < MX; i++) {
        if (!a[i]) {
            for (int j = i + i; j < MX; j += i) {
                a[j] = 1;
                if (!arr[j]) arr[j] = i;
            }
        }
    }
}
LL pow(LL a, LL b) {
    LL ret = 1;
    while (b) {
        if (b & 1) ret = ret * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ret;
}
LL f[MX];
int main() {
    LL t;
    int l, r;
    init();
    //freopen("in.txt", "r", stdin);
    scanf("%I64d%d%d", &t, &l, &r);
    for (int i = 2; i <= r; i++) {
        if (!arr[i]) f[i] = ((LL)i * (i - 1) % mod) * pow(2ll, mod - 2) % mod;
        else f[i] = (f[i / arr[i]] + f[arr[i]] * i / arr[i]) % mod;
    }
    LL ans = 0, k = 1;
    for (int i = l; i <= r; i++) {
        ans = (ans + k * f[i]) % mod;
        k = k * t % mod;
    }
    printf("%I64d\n", ans);
    return 0;
}