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;
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long int gcd(long long int a, long long int b) {
  if (a < b)
    return gcd(b, a);
  else if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int isPrime(long long int n) {
  long long int p = (long long int)sqrt(n);
  for (long long int i = (2); i <= (p); i += (1))
    if (n % i == 0) return 0;
  return 1;
}
long long int pow(long long int b, long long int e) {
  if (e == 0)
    return 1;
  else if (e % 2 == 0) {
    long long int a = pow(b, e / 2);
    return a * a;
  } else {
    long long int a = pow(b, e / 2);
    return b * a * a;
  }
}
long long int powm(long long int x, long long int y,
                   long long int m = 1000000007) {
  x = x % m;
  long long int res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= m;
    y = y >> 1;
    x = x * x;
    x %= m;
  }
  return res;
}
long long int modInverse(long long int a, long long int m) {
  return powm(a, m - 2, m);
}
long long int lt = 200;
vector<vector<long long int>> dp(1e5 + 5, vector<long long int>(lt + 1, -1));
long long int n;
vector<long long int> arr(1e5 + 5);
void solve() {
  cin >> n;
  for (long long int i = (0); i <= (n - 1); i += (1)) cin >> arr[i];
  for (long long int i = (n); i >= (1); i -= (1)) {
    for (long long int j = (1); j <= (min(n, lt)); j += (1)) {
      if (i + arr[i - 1] + j > n) {
        dp[i][j] = 1;
        continue;
      }
      dp[i][j] = 1 + dp[i + arr[i - 1] + j][j];
    }
  }
  long long int q;
  cin >> q;
  while (q--) {
    long long int p, k;
    cin >> p >> k;
    long long int val = p, steps = 0;
    if (k <= lt) {
      cout << dp[p][k] << '\n';
      continue;
    }
    while (val <= n) {
      val += arr[val - 1] + k;
      steps++;
    }
    cout << steps << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}
