include <bits/stdc++.h>

#pragma GCC optimize ('O3,unroll-loops') #pragma GCC target ("avx2") using namespace std;

define all(x) x.begin(),x.end()

define ll long long

define ld long double

define fuk return

define getunique(v) {sort(v.begin(), v.end()); v.erase(unique(v.begin(), v.end()), v.end());}

define pb push_back

define tr(it,a) for(auto it=a.begin();it!=a.end();it++)

define fo(i,n) for(int i=0;i<n;i++)

define fop(i,x,n) for(int i=x;i<=n;i++)

define forv(i,l,n) for(int i=l;i>=n;i--)

define nl << "\n";

typedef pair<ll,ll> pl; typedef vector vl; typedef vector < pair <ll,ll > > vp; typedef vector vb; typedef vector vd; typedef vector vs;

define inp(v, n) for(int i=0; i<n; ++i) cin >> v[i];

define opt(v) for(auto x: v) cout << x << ' '; cout << endl;

const ll mod = 1000000007; const ll N = 3e5+10;

define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

define int long long

ll binpow(ll a, ll b) { ll result = 1; while (b > 0) { if (b & 1) result *= a; a *= a; a %= mod; b /= 2; result %= mod; } return result; } ll n; ll a[N]; multiset mt; bool check(ll mid){ mt.clear(); fo(i,mid){ mt.insert(a[i]); } fop(i,mid,n){ ll mn=*mt.begin(); // set st; ll cnt=0;

// for(auto it=mt.begin();it!=mt.end();++it){
    //     st.insert(*it);
    // }
    for(ll j=mn;j<=1e7;j+=mn){
        // st.erase(st.find(j));
        cnt+=mt.count(j);
    }
    if(mt.size()==cnt){
        return true;
    }
    if(i!=n){
        mt.insert(a[i]);
        mt.erase(mt.find(a[i-mid]));
    }
}
return false;

} void solve(){

cin>>n;
fo(i,n) cin>>a[i];
ll mo=1e18;
fo(i,n){
    mo=min(mo,a[i]);
}
if(mo==1){
    cout<<"1 "<<n-1 nl
    cout<<"1\n";
    return;
}
ll lo=1;
ll hi=n;
ll ans=1;
//  cout<<"ho\n";
while(hi>=lo){
    ll mid=(lo+hi)/2;
    // cout<<"hi\n";
    // lo=hi+1;
    if(check(mid)){
        ans=mid;
        lo=mid+1;
    }else{
        hi=mid-1;
    }
}
ll value=0;
mt.clear();
fo(i,ans){
    mt.insert(a[i]);
}
vl ind;
fop(i,ans,n){
    ll mn=*mt.begin();
    // set<ll> st;
    ll cnt=0;

    // for(auto it=mt.begin();it!=mt.end();++it){
    //     st.insert(*it);
    // }
    for(ll j=mn;j<=1000004;j+=mn){
        // st.erase(st.find(j));
        cnt+=mt.count(j);
    }
    if(mt.size()==cnt){
        ind.push_back(i-ans+1);
        value++;
    }
    if(i!=n){
        mt.insert(a[i]);
        mt.erase(mt.find(a[i-ans]));
    }
}
sort(all(ind));
cout<<value<<" ";
cout<<ans-1 nl
fo(i,ind.size()){
    cout<<ind[i]<<" ";
}cout nl

} signed main(){ IOS #ifndef ONLINE_JUDGE freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); #endif solve(); return 0; }

~~~~~

Here I have used a simple binary search which is overall giving time complexity (o(n)logn) there is sieve technique used in code which can compute in O(nlog log n) time so overall the entire code wont take more than O(nlogn) as per me . Please Correct me if there is some mistake Question id is [https://codeforces.net/contest/359/problem/D] and the solution i submitted is [https://codeforces.net/contest/359/submission/227495328]