i am having a hard time in understanding the below code
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e5+5;
bool vis[N];
int a[N];
int n;
int s;
bool check(int x){
if(s%x!=0) return false;
for(int i=1; i<=n ;i++) vis[i]=false;
for(int i=1; i<=n ;i++){
if(a[i]%x==0  a[i]%x>n  vis[a[i]%x]) return false;
vis[a[i]%x]=true;
}
return true;
}
void solve()
{
cin>>n;
s = 0;
for(int i=1;i<=n;i++)
{
cin>>a[i];
s += (a[i]i);
}
if(s == 0) {
cout<<"YES "<<n+1<<"\n";
} else if(s < 0) {
cout<<"NO"<<"\n";
} else {
for (int i = 1; i * i <= s ; i++) {
if (s % i == 0) {
if(check(i) && i <= 2e7) {cout << "YES " << i << '\n';return;}
if(check(s/i) && (s/i) <= 2e7) {cout << "YES " << s/i << '\n';return;}
}
}
cout << "NO" << '\n';return;
}
}
int32_t main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
int t;
cin>>t;
while(t)
solve();
return 0;
}
Things i didn’t understood

why are we cheching the condition for s<0
doesn’t it means that Q’<0 (considering here s represents Q’) which is not possible? 
why answer is n+1 when s=0

why we have to check again in check function for the condition
s%x!=0
? isn’t it already handled in the for loop of solve function? 
the following part of check function
for(int i=1; i<=n ;i++){
if(a[i]%x==0  a[i]%x>n  vis[a[i]%x]) return false;
vis[a[i]%x]=true;
}