F. Ant colony

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mole is hungry again. He found one ant colony, consisting of n ants, ordered in a row. Each ant i (1 ≤ i ≤ n) has a strength si.

In order to make his dinner more interesting, Mole organizes a version of «Hunger Games» for the ants. He chooses two numbers l andr (1 ≤ l ≤ r ≤ n) and each pair of ants with indices between l and r (inclusively) will fight. When two ants i and j fight, ant i gets one battle point only if si divides sj (also, ant j gets one battle point only if sj divides si).

After all fights have been finished, Mole makes the ranking. An ant i, with vi battle points obtained, is going to be freed only if vi = r - l, or in other words only if it took a point in every fight it participated. After that, Mole eats the rest of the ants. Note that there can be many ants freed or even none.

In order to choose the best sequence, Mole gives you t segments [li, ri] and asks for each of them how many ants is he going to eat if those ants fight.

Input

The first line contains one integer n (1 ≤ n ≤ 105), the size of the ant colony.

The second line contains n integers s1, s2, ..., sn (1 ≤ si ≤ 109), the strengths of the ants.

The third line contains one integer t (1 ≤ t ≤ 105), the number of test cases.

Each of the next t lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n), describing one query.

Output

Print to the standard output t lines. The i-th line contains number of ants that Mole eats from the segment [li, ri].

Examples

input

5
1 3 2 4 2
4
1 5
2 5
3 5
4 5

output

4
4
1
1

Note

In the first test battle points for each ant are v = [4, 0, 2, 0, 2], so ant number 1 is freed. Mole eats the ants 2, 3, 4, 5.

In the second test case battle points are v = [0, 2, 0, 2], so no ant is freed and all of them are eaten by Mole.

In the third test case battle points are v = [2, 0, 2], so ants number 3 and 5 are freed. Mole eats only the ant 4.

In the fourth test case battle points are v = [0, 1], so ant number 5 is freed. Mole eats the ant 4.

Explanation

Let us begin with few observations.

(1) A mole cannot kill an ant if that ant divides every number in the range left to right

(2) Now infact the ant having the value equal to gcd of a[left...right] cannot be killed by the mole

Now we need to maintain a segment tree for gcd queries between left to right. And query the tree for gcd from left to right. Now all that is left is knowing how many such values lie between left to right.

This is cleverly done using a sorted pair of vector which stores values and indices. Now we do a lower bound query on the vector pair   using pair(value,right+1) and pair(value,left)  ,the difference of this indices gives us the number of values lying between left and right.

************ Code goes here*****************

#include<bits/stdc++.h>

using namespace std;

int a[200010];

int tree[600010];

#define ff first

#define ss second

#define mp make_pair

#define pb push_back

vector<pair<int,int> > v;

void build(int node,int s,int e)

{

  if(s>e)

  return ;

  if(s==e)

  {

      tree[node]=a[s];

      return ;

  }

  int mid=(s+e)/2;

  build(2*node,s,mid);

  build(2*node+1,mid+1,e);

  tree[node]=__gcd(tree[2*node],tree[2*node+1]);

  

}

int query(int node,int s,int e,int l,int r)

{

       if(s>e || s>r || e<l)

       return 0;

       if(s>=l && e<=r )

       return tree[node];

       int mid=(s+e)/2;

       int gc1=query(2*node,s,mid,l,r);

       int gc2=query(2*node+1,mid+1,e,l,r);

       return __gcd(gc1,gc2);

}

int get(int val,int idx)

{

       pair<int,int> mm;

       mm=make_pair(val,idx);

       int r=lower_bound(v.begin(),v.end(),mm)-v.begin();

       return r;

}

int main()

{

 int n;

 cin>>n;

 for(int i=0;i<n;i++)

 {

 

 cin>>a[i];

  v.pb(mp(a[i],i));

       }

       sort(v.begin(),v.end());

 build(1,0,n-1);

 int q;

 cin>>q;

 while(q--)

 {

    int l,r;

    scanf("%d %d",&l,&r);

    l--;

    r--;

    int gc=query(1,0,n-1,l,r);

    int hig=get(gc,r+1);

    int low=get(gc,l);

    int ans=(r-l+1)-(hig-low);

    printf("%d\n",ans);

    

 }

 return 0;

}