#include <bits/stdc++.h>
using namespace std;
int tata[1 << 19], ptr[1 << 19];
int a[1 << 19], ans[1 << 19];
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
for(int i = 1; i <= n; i ++)
cin >> a[i], a[i] += a[i - 1];
map<int, int> mp;
int last = n + 1;
mp[a[n]] = n;
ans[n + 2] = -1;
ans[n + 1] = 0;
tata[n + 1] = n + 2;
tata[n + 2] = n + 2;
for(int i = n; i; i --)
{
if(mp[a[i - 1]] != 0)
last = min(last, mp[a[i - 1]]);
mp[a[i - 1]] = i - 1;
tata[i] = last + 1;
ans[i] = 1 + ans[last + 1];
}
int q;
cin >> q;
for(int i = n + 2; i; i --)
{
if(ans[ptr[ptr[tata[i]]]] - ans[ptr[tata[i]]] == ans[ptr[tata[i]]] - ans[tata[i]])
ptr[i] = ptr[ptr[tata[i]]];
else
ptr[i] = tata[i];
}
for(int i = 1; i <= q; i ++)
{
int l, r, li;
cin >> l >> r;
li = l;
r ++;
while(1)
{
if(ptr[l] <= r)
l = ptr[l];
else if(tata[l] <= r)
l = tata[l];
else
break;
}
cout << ans[li] - ans[l] << '\n';
}
}
/*
ne trebuie prima pereche de sume partiale egale care se afla
la dreapta unui punct
si facem binary lifting pe tot
yessir NU AM MEMORIE CE E ASTA
am o idee
putem adauga incet incet lasturi de la dreapta la stanga
putem sa facem binary lifting in O(1) memorie ca la arbori
si adancimea va fi practic raspunsu?
ca e cazul ala
if(adanc[ptr[ptr[tata]]] - adanc[ptr[tata]] == adanc[ptr[tata]] - adanc[tata])
ptr[nod] = ptr[ptr[tata]];
si cam asta e tot sper
*/
/*
10
1 2 -3 0 1 -4 3 2 -1 1
3
1 10
1 5
2 9
*/
