#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
// Use slightly larger bounds to prevent off-by-one errors
const int MAXN = 200007;
const int LOGN = 20;
const int INF_IDX = 200005;
int jump[LOGN][MAXN];
ll pr[MAXN];
int a[MAXN];
void solve() {
int n;
if (!(cin >> n)) return;
// Use a standard map for safety; if it TLEs, we use a custom hash
map<ll, int> mp;
pr[0] = 0;
for(int i = 1; i <= n; ++i) {
cin >> a[i];
pr[i] = pr[i-1] + a[i];
}
// Initialize jump table with a value representing "out of bounds"
for(int j = 0; j < LOGN; j++) {
for(int i = 0; i <= n + 2; i++) {
jump[j][i] = n + 1;
}
}
int next_occurrence = n + 1;
mp[pr[n]] = n;
// Precomputing jump[0][i]: The first point where a zero-sum subarray ends
// starting from index i or later.
for (int i = n; i >= 1; --i) {
// Option 1: A zero-sum subarray already exists further to the right
next_occurrence = jump[0][i+1];
// Option 2: A zero-sum subarray ends right here (if a[i] == 0)
if (a[i] == 0) {
next_occurrence = min(next_occurrence, i);
}
// Option 3: A zero-sum subarray starts at i and ends at some k
// This happens if pr[k] == pr[i-1]
if (mp.find(pr[i-1]) != mp.end()) {
next_occurrence = min(next_occurrence, mp[pr[i-1]]);
}
jump[0][i] = next_occurrence;
mp[pr[i-1]] = i - 1; // Update map with the leftmost occurrence of this sum
}
// Binary Lifting Table Construction
for (int j = 1; j < LOGN; ++j) {
for (int i = 1; i <= n; ++i) {
int mid = jump[j-1][i];
if (mid < n) {
// If we found a zero-sum range ending at 'mid',
// the next one must start at 'mid + 1'
jump[j][i] = jump[j-1][mid + 1];
} else {
jump[j][i] = n + 1;
}
}
}
int q;
cin >> q;
while(q--) {
int l, r;
cin >> l >> r;
int cur = l;
int ans = 0;
for (int j = LOGN - 1; j >= 0; --j) {
// If the j-th jump starts at 'cur' and ends within 'r'
if (cur <= r && jump[j][cur] <= r) {
ans += (1 << j);
cur = jump[j][cur] + 1; // Next subarray must start after this one ends
}
}
cout << ans << '\n';
}
}
int main() {
// Fast I/O is essential for 2e5 inputs
ios::sync_with_stdio(false);
cin.tie(nullptr);
solve();
return 0;
}
