#include<cstdio>
#include<algorithm>
#define N 100001
using namespace std;
bool vis[N];
int n,tot,pri[N],d[N];
long long ans,f[N],ff[N];
int read()
{int x=0,f=1; char ch=getchar();while(ch<'0'||ch>'9') ch=getchar();while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}return x*f;
}
int work()
{ff[1]=1,f[1]=1;for(int i=2;i<=n;i++){if(!vis[i]){pri[tot++]=i;d[i]=1,ff[i]=2;}for(int j=0;j<tot&&i*pri[j]<=n;j++){vis[i*pri[j]]=true;if(i%pri[j]){ff[i*pri[j]]=ff[i]*2;d[i*pri[j]]=1;}else{ff[i*pri[j]]=ff[i]/(d[i]+1)*(d[i]+2);d[i*pri[j]]=d[i]+1;break;}} f[i]=f[i-1]+ff[i];}
}
int main()
{freopen("count.in","r",stdin);freopen("count.out","w",stdout);n=read();work();ans=f[n];printf("%I64d\n",ans);return 0;
}

#include<cstdio>
#include<algorithm>
#define N 10000100
using namespace std;
int n,tot,pri[N],d[N],ff[N];
long long ans;
int read()
{int x=0,f=1; char ch=getchar();while(ch<'0'||ch>'9') ch=getchar();while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}return x*f;
}
int work()
{ff[1]=1;ans=1;for(int i=2;i<=n;i++){if(!d[i]){pri[tot++]=i;d[i]=1,ff[i]=2;}for(int j=0;j<tot&&i*pri[j]<=n;j++){if(i%pri[j]){ff[i*pri[j]]=ff[i]*2;d[i*pri[j]]=1;}else{ff[i*pri[j]]=ff[i]/(d[i]+1)*(d[i]+2);d[i*pri[j]]=d[i]+1;break;}} ans+=ff[i];}
}
int main()
{freopen("count.in","r",stdin);freopen("count.out","w",stdout);n=read();work();printf("%I64d\n",ans);return 0;
}

#include<cstdio>
#include<algorithm>
using namespace std;
int n;
long long ans;
int read()
{int x=0,f=1; char ch=getchar();while(ch<'0'||ch>'9') ch=getchar();while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}return x*f;
}
int main()
{freopen("count.in","r",stdin);freopen("count.out","w",stdout);n=read();for(int i=1;i<=n;i++)ans+=n/i;printf("%I64d\n",ans);return 0;
}

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define N 1000
using namespace std;
bool vis[N];
long long ans=0x3f;
int n,m,x[N],f[N][N],dp[N][N],q[N];
int read()
{int x=0,f=1; char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1; ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}return x*f;
}
int main()
{freopen("post.in","r",stdin);freopen("post.out","w",stdout);n=read(),m=read();memset(dp,0x3f,sizeof(dp));for(int i=1;i<=n;i++)x[i]=read();for(int i=1;i<=n;i++)for(int j=i;j<=n;j++){int mid=(i+j)/2;for(int k=i;k<=j;k++)f[i][j]+=abs(x[k]-x[mid]);} //处理出只放一个邮局的情况。我们可以知道如果只让一个邮局的话，邮局的位置一定是放在中间的。 for(int i=1;i<=m;i++)dp[i][i]=0;//如果在当期点放邮局的话，那当前的最短距离为0 for(int i=1;i<=n;i++)//在i这个地方放邮局的话
    {dp[i][1]=f[1][i]; //放一个邮局且放在当前点的话最终值为从第一个点到当前点的便利度 for(int j=2;j<=min(m,i);j++)//枚举到当前点放几个邮局
        {if(i==j) continue;dp[i][j]=dp[i-1][j-1];for(int k=j-1;k<=i;k++)dp[i][j]=min(dp[i][j],dp[k][j-1]+f[k+1][i]);//枚举上一个放邮局的地方
        } }      ans=dp[n][m];printf("%I64d",ans);return 0;
}

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#define pz putchar('0');
#include <map>
using namespace std;
const int MO = 15;
struct Big{int len, data[10001];void clear() { memset(this, 0, sizeof(*this)); }int & operator [] (int k) { return data[k]; }Big & operator = (int k){clear();len = 0;while(k){++len; data[len] = k & MO; k >>= 4;} if (len == 0) ++len;return *this;}Big operator * (Big & A){Big temp; temp.clear(); temp.len = len + A.len - 1;for (int i = 1; i <= len; i++)for (int j = 1; j <= A.len; j++){temp[i + j - 1] += A[j] * data[i];temp[i + j] += (temp[i + j - 1] >> 4);temp[i + j - 1] &= MO;}while(temp[temp.len + 1]) ++temp.len;return temp;}void print(){for (int i = len; i >= 1; i--) printf("%X", data[i]);putchar('\n');}
} temp, ans;
map<int, bool> M;
bool f[1000001];
int pnum, p[100001];
void GETP(int M)
{memset(f, 1, sizeof(f)); f[0] = f[1] = false;p[pnum = 1] = 2;for (int now = 2; now < M;){for (int j = now + now; j <= M; j += now) f[j] = false;++now; while(now < M && !f[now]) ++now;if (f[now]) p[++pnum] = now;}
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
void work(int num)
{for (int i = 1; i <= pnum; i++){if (num % p[i] == 0)if (M[p[i]] == 0){M[p[i]] = true;temp = p[i];ans = ans * temp;}while(num % p[i] == 0) num /= p[i];}if (num != 1)if (M[num] == 0){M[num] = true;temp = num;ans = ans * temp;}
}
int main()
{freopen("fraction.in", "r", stdin); freopen("fraction.out", "w", stdout);ans = 1;int t; scanf("%d", &t);GETP(100000);while(t--){int a, b;scanf("%d%d", &a, &b);int d = gcd(a, b); a /= d; b /= d;work(b);}ans.print();
}