#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 100000 + 5; //MAXN = sqrt(MAXINPUTN)
const ll MOD = 1000000007, inv2 = 500000004, inv3 = 333333336;
int num;
const int MAXP = 10000; //MAXN以内质数的数量,至多与MAXN一样大
ll prime[MAXP], sp1[MAXP], sp2[MAXP]; //在调用init1()之后,可以精确计算出这里需要的大小
int tot;
const int DMAXN = (MAXN<<1) + 5; //整除分块需要的空间
ll n, SQRT, g1[DMAXN], g2[DMAXN], w[DMAXN], ind1[DMAXN], ind2[DMAXN];
void init1() { //线性筛预处理
SQRT = sqrt(n), num = 0 ;
bitset < MAXN+5 > notprime;
notprime[1] = 1;
for(int i = 2; i <= SQRT; ++i) {
if(!notprime[i]) {
prime[++num] = i;
sp1[num] = (sp1[num - 1] + i) % MOD;
sp2[num] = (sp2[num - 1] + 1ll * i * i) % MOD;
//spx[num] = 前num个质数的 x 次方和
}
for(int j = 1; j <= num && prime[j]*i <= SQRT; ++j) {
notprime[i * prime[j]] = 1;
if(i % prime[j] == 0)
break;
}
}
printf("prime num=%d
", num); //传入最大的n,算出num之后给MAXP赋值
}
void init2() {
tot = 0;
for(ll l = 1, r, t; l <= n; l = r + 1) {
t = n / l, r = n / t;
w[++tot] = t;
g1[tot] = w[tot] % MOD;
g2[tot] = (((g1[tot] * (g1[tot] + 1)) >> 1) % MOD * (2 * g1[tot] + 1) % MOD * inv3 % MOD) - 1;
g1[tot] = ((g1[tot] * (g1[tot] + 1)) >> 1) % MOD - 1;
if(t <= SQRT)
ind1[t] = tot;
else
ind2[r] = tot;
}
//gx(n,j) 表示 [1,n]中,i是质数或者i的最小质因子>pj的数的 x 次方和,n滚动省略
//ind1和ind2用来记录这个数在数组中的位置
for(int i = 1; i <= num; i++) {
for(int j = 1; j <= tot && prime[i]*prime[i] <= w[j]; j++) {
ll k = w[j] / prime[i] <= SQRT ? ind1[w[j] / prime[i]] : ind2[n / (w[j] / prime[i])];
g1[j] -= prime[i] * (g1[k] - sp1[i - 1] + MOD) % MOD;
g2[j] -= prime[i] * prime[i] % MOD * (g2[k] - sp2[i - 1] + MOD) % MOD;
g1[j] %= MOD, g2[j] %= MOD;
if(g1[j] < 0)
g1[j] += MOD;
if(g2[j] < 0)
g2[j] += MOD;
}
}
}
ll S(ll x, int y) { //第二部分,不要记忆化,记忆化又慢又卡
if(prime[y] >= x)
return 0;
int k = (x <= SQRT) ? ind1[x] : ind2[n / x];
ll ans = (g2[k] - g1[k] + MOD - (sp2[y] - sp1[y]) + MOD) % MOD;
for(int i = y + 1; i <= num && prime[i]*prime[i] <= x; i++) {
ll pe = prime[i];
for(int e = 1; pe <= x; ++e, pe = pe * prime[i]) {
ll xx = pe % MOD;
ans = (ans + xx * (xx - 1) % MOD * (S(x / pe, i) + (e != 1))) % MOD;
}
}
return ans;
}
int main() {
#ifdef Yinku
freopen("Yinku.in", "r", stdin);
#endif // Yinku
scanf("%lld", &n);
init1();
init2();
printf("%lld
", (S(n, 0) + 1) % MOD); //加上f(1)
return 0;
}