Consider a sequence of digits of length 2^k [a_1, a_2, …, a_{2^k}]. We perform the following operation with it: replace pairs (a_{2i+1}, a_{2i+2}) with (a_{2i+1} + a_{2i+2})mod 10 for 0≤ i<2^{k-1}. For every i where a_{2i+1} + a_{2i+2}≥ 10 we get a candy! As a result, we will get a sequence of length 2^{k-1}.

Less formally, we partition sequence of length 2^k into 2^{k-1} pairs, each consisting of 2 numbers: the first pair consists of the first and second numbers, the second of the third and fourth …, the last pair consists of the (2^k-1)-th and (2^k)-th numbers. For every pair such that sum of numbers in it is at least 10, we get a candy. After that, we replace every pair of numbers with a remainder of the division of their sum by 10 (and don't change the order of the numbers).

Perform this operation with a resulting array until it becomes of length 1. Let f([a_1, a_2, …, a_{2^k}]) denote the number of candies we get in this process. 

For example: if the starting sequence is [8, 7, 3, 1, 7, 0, 9, 4] then:

After the first operation the sequence becomes [(8 + 7)mod 10, (3 + 1)mod 10, (7 + 0)mod 10, (9 + 4)mod 10] = [5, 4, 7, 3], and we get 2 candies as 8 + 7 ≥ 10 and 9 + 4 ≥ 10.

After the second operation the sequence becomes [(5 + 4)mod 10, (7 + 3)mod 10] = [9, 0], and we get one more candy as 7 + 3 ≥ 10. 

After the final operation sequence becomes [(9 + 0) mod 10] = [9]. 

Therefore, f([8, 7, 3, 1, 7, 0, 9, 4]) = 3 as we got 3 candies in total.

You are given a sequence of digits of length n s_1, s_2, … s_n. You have to answer q queries of the form (l_i, r_i), where for i-th query you have to output f([s_{l_i}, s_{l_i+1}, …, s_{r_i}]). It is guaranteed that r_i-l_i+1 is of form 2^k for some nonnegative integer k.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the sequence.

The second line contains n digits s_1, s_2, …, s_n (0 ≤ s_i ≤ 9).

The third line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.

Each of the next q lines contains two integers l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — i-th query. It is guaranteed that r_i-l_i+1 is a nonnegative integer power of 2.

Output

Output q lines, in i-th line output single integer — f([s_{l_i}, s_{l_i + 1}, …, s_{r_i}]), answer to the i-th query.

Examples

Input


8
8 7 3 1 7 0 9 4
3
1 8
2 5
7 7


Output


3
1
0


Input


6
0 1 2 3 3 5
3
1 2
1 4
3 6


Output


0
0
1

Note

The first example illustrates an example from the statement.

f([7, 3, 1, 7]) = 1: sequence of operations is [7, 3, 1, 7] → [(7 + 3)mod 10, (1 + 7)mod 10] = [0, 8] and one candy as 7 + 3 ≥ 10 → [(0 + 8) mod 10] = [8], so we get only 1 candy.

f([9]) = 0 as we don't perform operations with it.
#include <bits/stdc++.h>
using namespace std;
struct abc {
  int p, q, idx;
};
bool compareTwoStudents(abc a, abc b) {
  if (a.p != b.p) return a.p < b.p;
  return (a.q < b.q);
}
int getMid(int s, int e) { return s + (e - s) / 2; }
int getSumUtil(int *st, int ss, int se, int qs, int qe, int si) {
  if (qs <= ss && qe >= se) return st[si];
  if (se < qs || ss > qe) return 0;
  int mid = getMid(ss, se);
  return getSumUtil(st, ss, mid, qs, qe, 2 * si + 1) +
         getSumUtil(st, mid + 1, se, qs, qe, 2 * si + 2);
}
void updateValueUtil(int *st, int ss, int se, int i, int diff, int si) {
  if (i < ss || i > se) return;
  st[si] = st[si] + diff;
  if (se != ss) {
    int mid = getMid(ss, se);
    updateValueUtil(st, ss, mid, i, diff, 2 * si + 1);
    updateValueUtil(st, mid + 1, se, i, diff, 2 * si + 2);
  }
}
void updateValue(int arr[], int *st, int n, int i, int new_val) {
  if (i < 0 || i > n - 1) {
    cout << "Invalid Input";
    return;
  }
  int diff = new_val - arr[i];
  arr[i] = new_val;
  updateValueUtil(st, 0, n - 1, i, diff, 0);
}
int getSum(int *st, int n, int qs, int qe) {
  if (qs < 0 || qe > n - 1 || qs > qe) {
    cout << "Invalid Input";
    return -1;
  }
  return getSumUtil(st, 0, n - 1, qs, qe, 0);
}
int constructSTUtil(int arr[], int ss, int se, int *st, int si) {
  if (ss == se) {
    st[si] = arr[ss];
    return arr[ss];
  }
  int mid = getMid(ss, se);
  st[si] = constructSTUtil(arr, ss, mid, st, si * 2 + 1) +
           constructSTUtil(arr, mid + 1, se, st, si * 2 + 2);
  return st[si];
}
int *constructST(int arr[], int n) {
  int x = (int)(ceil(log2(n)));
  int max_size = 2 * (int)pow(2, x) - 1;
  int *st = new int[max_size];
  constructSTUtil(arr, 0, n - 1, st, 0);
  return st;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int i, j, k, m, n, t, u, v, w, p, q;
  long long int x, y, z;
  string s, s1, s2;
  vector<int> abc;
  vector<int>::iterator it;
  priority_queue<int> pq;
  map<char, int> pm;
  cin >> n;
  int arr[n - 1];
  for (int i = 0; i <= n - 1; ++i) {
    cin >> arr[i];
  }
  int *st = constructST(arr, n);
  cin >> q;
  while (q--) {
    cin >> u >> v;
    u--;
    v--;
    int w = getSum(st, n, u, v);
    cout << w / 10 << "\n";
  }
  return 0;
}
