Problem: This time Baby Ehab will only cut and not stick. He starts with a piece of paper with an array a of length n written on it, and then he does the following:

  * he picks a range (l, r) and cuts the subsegment a_l, a_{l + 1}, …, a_r out, removing the rest of the array. 
  * he then cuts this range into multiple subranges. 
  * to add a number theory spice to it, he requires that the elements of every subrange must have their product equal to their [least common multiple (LCM)](https://en.wikipedia.org/wiki/Least_common_multiple). 



Formally, he partitions the elements of a_l, a_{l + 1}, …, a_r into contiguous subarrays such that the product of every subarray is equal to its LCM. Now, for q independent ranges (l, r), tell Baby Ehab the minimum number of subarrays he needs.

Input

The first line contains 2 integers n and q (1 ≤ n,q ≤ 10^5) — the length of the array a and the number of queries.

The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^5) — the elements of the array a.

Each of the next q lines contains 2 integers l and r (1 ≤ l ≤ r ≤ n) — the endpoints of this query's interval.

Output

For each query, print its answer on a new line.

Example

Input


6 3
2 3 10 7 5 14
1 6
2 4
3 5


Output


3
1
2

Note

The first query asks about the whole array. You can partition it into [2], [3,10,7], and [5,14]. The first subrange has product and LCM equal to 2. The second has product and LCM equal to 210. And the third has product and LCM equal to 70. Another possible partitioning is [2,3], [10,7], and [5,14].

The second query asks about the range (2,4). Its product is equal to its LCM, so you don't need to partition it further.

The last query asks about the range (3,5). You can partition it into [10,7] and [5].

Solution (in CPP):

Solution:


//Shrey Dubey


#include<iostream>
#include<string>
#include<algorithm>
#include<map>
#include<unordered_map>
#include<vector>
#include<set>
#include<list>
#include<iomanip>
#include<queue>
#include<stack>
#include <math.h>
#include<climits>
#include<bitset>
#include<cstring>
#include<numeric>
#include<array>


using namespace std;
typedef long long ll;
typedef long double ld;

#define YES cout<<"YES\n"
#define Yes cout<<"Yes\n"
#define NO cout<<"NO\n"
#define No cout<<"No\n"
#define prDouble(x) cout<<fixed<<setprecision(10)<<x //to print decimal numbers
#define pb push_back
#define ff first
#define sec second
#define umap unordered_map
#define mp make_pair
#define KOBE ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define fo(n) for(ll i = 0; i<n; i++)
#define fnd(stl, data) find(stl.begin(), stl.end(), data)
#define forn(x,n) for(ll x = 0; x<n; x++)
#define imax INT_MAX
#define lmax LLONG_MAX
#define imin INT_MIN
#define lmin LLONG_MIN
#define vi vector<int>
#define vl vector<ll>
#define vp vector<pair<ll,ll> >
#define vs vector<string>
#define vb vector<bool>
#define pr(t) cout<<t<<"\n"
#define int long long
#define ql queue<ll>
#define qp queue<pair<ll,ll> >
#define endl "\n"
#define nl cout<<"\n"
#define re cin >>
#define pll pair<ll,ll>
#define FOR(a,b) for(ll i = a; i<=b; i++)
#define all(x) x.begin(),x.end()
#define LG 22

// ll dx[] = {1,0,-1,0};
// ll dy[] = {0,1,0,-1};

ll mod = 1e9 + 7;

ll cl(ld a){
    if(a>(ll) a){
        return (ll)a+1;
    }
    else{
        return (ll)a;
    }
}

ll flr(ld a){
    if(a < 0.0){
        return (ll) a - 1;
    }
    return (ll) a;
}

ll gcd(ll a, ll b){
    if(b == 0) return a;
    else return gcd(b,a%b);
}

//code starts here

ll n,q;
const ll M = 1e5+100;
ll nxt[M] = {0}, a[M];
ll dp[LG][M] = {0};
vl fact[M];
void solve(){
    re n; re q;
    fo(n) re a[i+1];
    for(ll i = 2; i<M; i++){
        if(!fact[i].empty()) continue;
        for(ll j = i; j<M; j+=i){
            nxt[i] = n+1;
            fact[j].pb(i);
        }
    }
    dp[0][n+1] = n+1;
    for(ll i = n; i>=1; i--){
        dp[0][i] = dp[0][i+1];
        for(auto f: fact[a[i]]){
            // if(i == 6 && f == 7) pr("here");
            dp[0][i] = min(dp[0][i],nxt[f]);
            nxt[f] = i;
        }
        // cout<<i<<"()"<<dp[0][i]<<"\n";
    }

    for(ll j = 1; j<20; j++){
        for(ll i = 1; i<=n+1; i++){
            ll k = dp[j-1][i];
            dp[j][i] = dp[j-1][k];
        }
    }
    while(q--){
        ll l,r; re l; re r;
        ll ans = 0;
        for(ll i = 19; i>=0; i--){
            if(dp[i][l] <= r){
                // cout<<l<<"()\n";
                ans += (1<<i);
                l = dp[i][l];
            }
        }
        pr(ans+1);
    }
}

int32_t main(){
    KOBE;
    ll t;
    t = 1;
    // re t;
    while(t--) solve();
}


//common errors
// row - n, col - m always and loop var
// see the freq of numbers carefully
// see if there's array overflow
// use map for large inputs
// always check for n = 1 condition
// check loop starting and end points :(

//problem ideas
//check piegonhole wherever possible
//there might be many instances of limited answers like 0,1,2 only
// see suffix and prefix
//don't be obsessed with binary search
// try to find repeating pattern in matrices