/*

CodeForces 839D - Winter is here [ 数论,容斥 ] | Codeforces Round #428 (Div. 2)

题意：

	给出数列a[N]

	对每个子集，若 gcd(a[I1], a[I2], a[I3] ..., a[In]) > 1，则贡献为 n*gcd

	求总贡献和

	限制： N <= 2e5,a[i] <= 1e6

分析：

	记录 num[i]数组为 i 的倍数的个数

	则 gcd >= i 能组成的所有方案的总人数 f(i) = 2^(num[i]-1)*num[i]

	设 g(i) 为 gcd == i 能组成的所有方案的总人数

	可得 f(x) = ∑ [x|y] g(y)

	反演或者容斥即可

*/

#include <bits/stdc++.h>

using namespace std;

#define LL long long

const int N = 1e6+5;

const LL MOD = 1e9+7;

int n, a[N], num[N], Max;

LL two[N], sum[N];

int main()

{

    two[0] = 1;

    for (int i = 1; i < N; i++) two[i] = two[i-1] * 2 % MOD;

    scanf("%d", &n);

    Max = 0;

    for (int i = 1; i <= n; i++)

    {

        scanf("%d", &a[i]);

        Max = max(a[i], Max);

        num[a[i]]++;

    }

    for (int i = 1; i <= Max; i++)

        for (int j = i+i; j <= Max; j += i)

            num[i] += num[j];

    LL ans = 0;

    for (int i = Max; i >= 2; i--)

    {

        sum[i] = two[num[i]-1]*num[i] % MOD;

        for (int j = i+i; j <= Max; j += i)

        {

            sum[i] = (sum[i] - sum[j] + MOD) % MOD;

        }

        ans = (ans + sum[i] * i % MOD) % MOD;

    }

    printf("%lld\n", ans);

}

#include <bits/stdc++.h>

using namespace std;

#define LL long long

const LL MOD = 1e9+7;

const int N = 1000005;

bool notp[N];

int prime[N], pnum, mu[N];

void Mobius() {

	memset(notp, 0, sizeof(notp));

	mu[1] = 1;

	for (int i = 2; i < N; i++) {

		if (!notp[i]) prime[++pnum] = i, mu[i] = -1;

		for (int j = 1; prime[j]*i < N; j++) {

			notp[prime[j]*i] = 1;

			if (i%prime[j] == 0) {

				mu[prime[j]*i] = 0;

				break;

			}

			mu[prime[j]*i] = -mu[i];

		}

	}

}

int n, a[N], Max;

int num[N];

LL two[N];

int main()

{

    two[0] = 1;

    for (int i = 1; i < N; i++) two[i] = two[i-1]*2 % MOD;

    Mobius();

    scanf("%d", &n);

    Max = 0;

    for (int i = 1; i <= n; i++)

    {

        scanf("%d", &a[i]);

        Max = max(Max, a[i]);

        for (LL j = 1; j*j <= a[i]; j++)

        {

            if (j*j == a[i]) num[j]++;

            else if (a[i] % j == 0)

                num[j]++, num[a[i]/j]++;

        }

    }

    LL ans = 0;

    for (int i = 2; i <= Max; i++)

    {

        LL sum = 0;

        for (int j = i, k = 1; j <= Max; j += i, k++)

        {

            sum += (mu[k] * (two[num[j]-1]*num[j])%MOD + MOD) % MOD;

            sum %= MOD;

        }

        ans = (ans + sum * i%MOD) % MOD;

    }

    printf("%lld\n", ans% MOD);

}