#include #include #include #include #include #include #include #include #include using namespace std; #define int long long #define double long double #define rep(s,i,n) for(int i=s;i>n; #define sc(s) string s;cin>>s; #define dc(d) double d;cin>>d; #define mod 1000000007 #define inf 1000000000000000007 #define f first #define s second #define mini(c,a,b) *min_element(c+a,c+b) #define maxi(c,a,b) *max_element(c+a,c+b) #define pi 3.141592653589793238462643383279 #define e_ 2.718281828459045235360287471352 #define P pair #define upp(a,n,x) upper_bound(a,a+n,x)-a; #define low(a,n,x) lower_bound(a,a+n,x)-a; #define UF UnionFind //printf("%.12Lf\n",); int keta(int x) { rep(0, i, 30) { if (x < 10) { return i + 1; } x = x / 10; } } int gcd(int x, int y) { if (x == 0 || y == 0)return x + y; int aa = x, bb = y; rep(0, i, 1000) { aa = aa % bb; if (aa == 0) { return bb; } bb = bb % aa; if (bb == 0) { return aa; } } } int lcm(int x, int y) { int aa = x, bb = y; rep(0, i, 1000) { aa = aa % bb; if (aa == 0) { return x / bb * y; } bb = bb % aa; if (bb == 0) { return x / aa * y; } } } int integer(double d){ return long(d); } int distance(double a,double b,double c,double d){ return sqrt((b-a)*(b-a)+(c-d)*(c-d)); } bool prime(int x) { if (x == 1)return false; rep(2, i, sqrt(x) + 1) { if (x % i == 0 && x != i) { return false; } } return true; } int max(int a, int b) { if (a >= b)return a; else return b; } string maxst(string s, string t) { int n = s.size(); int m = t.size(); if (n > m)return s; else if (n < m)return t; else { rep(0, i, n) { if (s[i] > t[i])return s; if (s[i] < t[i])return t; } return s; } } int min(int a, int b) { if (a >= b)return b; else return a; } int n2[60]; int nis[60]; int nia[60]; int mody[60]; int nn; int com(int n, int y) { int ni = 1; for (int i = 0;i < 59;i++) { n2[i] = ni; ni *= 2; } int bunsi = 1, bunbo = 1; rep(0, i, y)bunsi = (bunsi * (n - i)) % mod; rep(0, i, y)bunbo = (bunbo * (i + 1)) % mod; mody[0] = bunbo; rep(1, i, 59) { bunbo = (bunbo * bunbo) % mod; mody[i] = bunbo; } rep(0, i, 59)nis[i] = 0; nn = mod - 2; for (int i = 58;i >= 0;i -= 1) { if (nn > n2[i]) { nis[i]++; nn -= n2[i]; } } nis[0]++; rep(0, i, 59) { if (nis[i] == 1) { bunsi = (bunsi * mody[i]) % mod; } } return bunsi; } int newcom(int n,int y){ int bunsi = 1, bunbo = 1; rep(0, i, y){ bunsi = (bunsi * (n - i)) ; bunbo = (bunbo * (i + 1)) ; int k=gcd(bunsi,bunbo); bunsi/=k; bunbo/=k; } return bunsi/bunbo; } int gyakugen(int n, int y) { int ni = 1; for (int i = 0;i < 59;i++) { n2[i] = ni; ni *= 2; } mody[0] = y; rep(1, i, 59) { y = (y * y) % mod; mody[i] = y; } rep(0, i, 59)nis[i] = 0; nn = mod - 2; for (int i = 58;i >= 0;i -= 1) { if (nn > n2[i]) { nis[i]++; nn -= n2[i]; } } nis[0]++; rep(0, i, 59) { if (nis[i] == 1) { n = (n * mody[i]) % mod; } } return n; } int yakuwa(int n) { int sum = 0; rep(1, i, sqrt(n + 1)) { if (n % i == 0)sum += i + n / i; if (i * i == n)sum -= i; } return sum; } int poow(int y, int n) { if (n == 0)return 1; n -= 1; int ni = 1; for (int i = 0;i < 58;i++) { n2[i] = ni; ni *= 2; } int yy = y; mody[0] = yy; rep(1, i, 58) { yy = (yy * yy) % mod; mody[i] = yy; } rep(0, i, 58)nis[i] = 0; nn = n; for (int i = 57;i >= 0;i -= 1) { if (nn >= n2[i]) { nis[i]++; nn -= n2[i]; } } rep(0, i, 58) { if (nis[i] == 1) { y = (y * mody[i]) % mod; } } return y; } int minpow(int x, int y) { int sum = 1; rep(0, i, y)sum *= x; return sum; } int ketawa(int x, int sinsuu) { int sum = 0; rep(0, i, 100)sum += (x % poow(sinsuu, i + 1)) / (poow(sinsuu, i)); return sum; } int sankaku(int a) { if(a%2==0) return a /2*(a+1); else return (a+1)/2*a; } int sames(int a[1111111], int n) { int ans = 0; rep(0, i, n) { if (a[i] == a[i + 1]) { int j = i; while (a[j + 1] == a[i] && j <= n - 2)j++; ans += sankaku(j - i); i = j; } } return ans; } using Graph = vector>; int oya[114514]; int depth[114514]; void dfs(const Graph& G, int v, int p, int d) { depth[v] = d; oya[v] = p; for (auto nv : G[v]) { if (nv == p) continue; // nv が親 p だったらダメ dfs(G, nv, v, d + 1); // d を 1 増やして子ノードへ } } /*int H=10,W=10; char field[10][10]; char memo[10][10]; void dfs(int h, int w) { memo[h][w] = 'x'; // 八方向を探索 for (int dh = -1; dh <= 1; ++dh) { for (int dw = -1; dw <= 1; ++dw) { if(abs(0-dh)+abs(0-dw)==2)continue; int nh = h + dh, nw = w + dw; // 場外アウトしたり、0 だったりはスルー if (nh < 0 || nh >= H || nw < 0 || nw >= W) continue; if (memo[nh][nw] == 'x') continue; // 再帰的に探索 dfs(nh, nw); } } }*/ int XOR(int a, int b) { if (a == 0 || b == 0) { return a + b; } int ni = 1; rep(0, i, 41) { n2[i] = ni; ni *= 2; } rep(0, i, 41)nis[i] = 0; for (int i = 40;i >= 0;i -= 1) { if (a >= n2[i]) { nis[i]++; a -= n2[i]; } if (b >= n2[i]) { nis[i]++; b -= n2[i]; } } int sum = 0; rep(0, i, 41)sum += (nis[i] % 2 * n2[i]); return sum; } //int ma[1024577][21]; //for(int bit=0;bit<(1< par; // par[i]:iの親の番号 (例) par[3] = 2 : 3の親が2 UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化 for (int i = 0; i < N; i++) par[i] = i; } int root(int x) { // データxが属する木の根を再帰で得る:root(x) = {xの木の根} if (par[x] == x) return x; return par[x] = root(par[x]); } void unite(int x, int y) { // xとyの木を併合 int rx = root(x); //xの根をrx int ry = root(y); //yの根をry if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時:xの根rxをyの根ryにつける } bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す int rx = root(x); int ry = root(y); return rx == ry; } }; signed main(){ ic(n) int sum=0; int s; rep(0,i,10000000){ if(i*i>n){ s=i; break; } } rep(1,i,s-1){ int l=n/i,r=n/(i+1)+1; int a=n%l,b=n%r; sum+=(a+b)*(l-r+1)/2; sum+=i*(l-r+1); sum%=mod; } int t=2; while(t*t<=n){ int l=1; int cnt=1; while(l<=n){ l*=t; cnt++; } int k=1; rep(0,i,cnt){ sum+=(n%k*t)/k; k*=t; } sum%=mod; t++; } c(sum) }