//枚举因子，查找和i最近的左右是i因子的点即可。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define LL long long
using namespace std;const int MAX=100010;
const LL mod=1e9+7;
int l_next[10010];
int r_next[10010];
int num[MAX];
int l[MAX],r[MAX];int main(){int n;while(scanf("%d",&n)!=EOF){for(int i=1;i<=n;i++){scanf("%d",&num[i]);l[i]=0;r[i]=MAX;}for(int i=0;i<10010;i++){l_next[i]=0; r_next[i]=MAX;}for(int i=1;i<=n;i++){int lmax=0;for(int k=1;k*k<=num[i];k++){if(num[i]%k==0){lmax=max(lmax,l_next[k]);// if(k!=1)lmax=max(lmax,l_next[num[i]/k]);}}l[i]=lmax;l_next[num[i]]=i;}/* for(int i=1;i<=n;i++)cout<<l[i]<<" ";cout <<endl;*/for(int i=n;i>=1;i--){int rmin=n+1;for(int k=1;k*k<=num[i];k++){if(num[i]%k==0){rmin=min(rmin,r_next[k]);// if(k!=1)rmin=min(rmin,r_next[num[i]/k]);}}r[i]=rmin;r_next[num[i]]=i;}/* for(int i=1;i<=n;i++)cout<<r[i]<<" ";cout <<endl;*/LL ans=0;for(int i=1;i<=n;i++){ans+=(LL)(i-l[i])*(LL)(r[i]-i);ans%=mod;}printf("%lld\n",ans);}return 0;
}
)
