a is an array of n positive integers, all of which are not greater than n.

You have to process q queries to this array. Each query is represented by two numbers p and k. Several operations are performed in each query; each operation changes p to p + ap + k. There operations are applied until p becomes greater than n. The answer to the query is the number of performed operations.

Input

The first line contains one integer n (1 ≤ n ≤ 100000).

The second line contains n integers — elements of a (1 ≤ ai ≤ n for each i from 1 to n).

The third line containts one integer q (1 ≤ q ≤ 100000).

Then q lines follow. Each line contains the values of p and k for corresponding query (1 ≤ p, k ≤ n).

Output

Print q integers, ith integer must be equal to the answer to ith query.

Example

Input

3
1 1 1
3
1 1
2 1
3 1


Output

2
1
1

Note

Consider first example:

In first query after first operation p = 3, after second operation p = 5.

In next two queries p is greater than n after the first operation.
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void unique(T &c) {
  c.erase(std::unique(c.begin(), c.end()), c.end());
}
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
  for (auto &v : vec) is >> v;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
  for (int i = 0; i < vec.size(); i++) {
    os << vec[i];
    if (i + 1 != vec.size()) os << " ";
  }
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &st) {
  for (auto itr = st.begin(); itr != st.end(); ++itr) {
    os << *itr;
    auto titr = itr;
    if (++titr != st.end()) os << " ";
  }
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_set<T> &st) {
  for (auto itr = st.begin(); itr != st.end(); ++itr) {
    os << *itr;
    auto titr = itr;
    if (++titr != st.end()) os << " ";
  }
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &st) {
  for (auto itr = st.begin(); itr != st.end(); ++itr) {
    os << *itr;
    auto titr = itr;
    if (++titr != st.end()) os << " ";
  }
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_multiset<T> &st) {
  for (auto itr = st.begin(); itr != st.end(); ++itr) {
    os << *itr;
    auto titr = itr;
    if (++titr != st.end()) os << " ";
  }
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const map<T1, T2> &mp) {
  for (auto itr = mp.begin(); itr != mp.end(); ++itr) {
    os << "(" << itr->first << ", " << itr->second << ")";
    auto titr = itr;
    if (++titr != mp.end()) os << " ";
  }
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp) {
  for (auto itr = mp.begin(); itr != mp.end(); ++itr) {
    os << "(" << itr->first << ", " << itr->second << ")";
    auto titr = itr;
    if (++titr != mp.end()) os << " ";
  }
  return os;
}
using ll = long long int;
using P = pair<int, int>;
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
struct fast_io {
  fast_io() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(20);
  }
} fast_io_;
const int B = 300;
int cnt[100010];
int p[100010], k[100010];
vector<int> ks[100010];
int res[100010];
int main(int argc, char const *argv[]) {
  int n;
  cin >> n;
  vector<int> a(n);
  cin >> a;
  int q;
  cin >> q;
  for (int i = (0); i < (q); ++i) {
    cin >> p[i] >> k[i];
    ks[k[i]].emplace_back(i);
  }
  for (int i = (1); i < (B + 1); ++i) {
    if (((int)(ks[i]).size()) == 0) continue;
    for (int j = n; j > 0; j--) {
      if (j + a[j - 1] + i > n)
        cnt[j] = 1;
      else
        cnt[j] = 1 + cnt[j + a[j - 1] + i];
    }
    for (auto idx : ks[i]) {
      res[idx] = cnt[p[idx]];
    }
  }
  for (int i = (0); i < (q); ++i) {
    if (k[i] <= B)
      cout << res[i] << "\n";
    else {
      int ans = 0;
      int curr = p[i];
      while (curr <= n) {
        ans++;
        curr = curr + a[curr - 1] + k[i];
      }
      cout << ans << "\n";
    }
  }
  cout << flush;
  return 0;
}
