for(int i=1;i<=n;i++){//1 indexed array

ST[i][0]=arr[i];//ST[i][j] will have ans for range[i,i+2^j-1]

}

for(int j=1;j<L;j++){

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

ST[i][j]=min(ST[i][j-1],ST[i+(1<<(j-1))][j-1]);

}

}

int ans=INT_MAX;

int dis=r-l+1;

//now the tricky part used in case of idempotent functions

int lr=log2(dis);

int ld=1<<lr;

// ans=min(ST[l][lr],ST[r-ld+1][lr]);

//another trick using C++ builting bit manipulation

int ln=31-b_clz(r-l+1);//no of leading zeros

ans=min(ST[l][ln],ST[r-ln+1][ln]);

return ans;

for(int i=1;i<=n;i++){//1 indexed array

ST[i][0]=arr[i];//ST[i][j] will have ans for range[i,i+2^j-1]

}

for(int j=1;j<L;j++){

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

ST[i][j]=__gcd(ST[i][j-1],ST[i+(1<<(j-1))][j-1]);

}

}

}