#include <bits/stdc++.h>
using namespace std;
const long double pi = 3.14159265358979;
const long long int mod = 1e9 + 7;
long long int takemod(long long int a, long long int mod) {
  return ((a % mod) + mod) % mod;
}
long long int bs(long long int a[], long long int l, long long int r,
                 long long int x) {
  if (r >= l) {
    long long int mid = (l + r) / 2;
    if (a[mid] == x)
      return mid;
    else if (a[mid] > x)
      return bs(a, l, mid - 1, x);
    else
      return bs(a, mid + 1, r, x);
  }
  return -1;
}
long long int s_t_i(string second) {
  stringstream geek(second);
  long long int m;
  geek >> m;
  return m;
}
long long int power(long long int b, long long int p) {
  if (p == 0)
    return 1;
  else if (p % 2 == 0)
    return (power(b, p / 2) * (power(b, p / 2)));
  else
    return ((b) * (power(b, p - 1)));
}
bool isprime(long long int n) {
  if (n == 1) return false;
  if (n == 2 or n == 3) return true;
  long long int lim = sqrt(n);
  long long int i = 2;
  while (i <= lim) {
    if (n % i == 0) return false;
    i++;
  }
  return true;
}
vector<long long int> factors(long long int n) {
  vector<long long int> a;
  a.push_back(1);
  long long int lim = sqrt(n), i = 2;
  while (i <= lim) {
    if (n % i == 0) {
      a.push_back(i);
      if (i != n / i) a.push_back(n / i);
    }
    i++;
  }
  if (n != 1) a.push_back(n);
  sort(a.begin(), a.end());
  return a;
}
vector<pair<long long int, long long int> > pf_with_count(long long int b) {
  pair<long long int, long long int> p;
  vector<pair<long long int, long long int> > pf;
  long long int count = 0;
  while (b % 2 == 0) {
    b /= 2;
    count++;
  }
  if (count > 0) {
    p.first = 2;
    p.second = count;
    pf.push_back(p);
  }
  long long int i = 3;
  while (i <= sqrt(b)) {
    count = 0;
    while (b % i == 0) {
      b /= i;
      count++;
    }
    if (count > 0) {
      p.first = i;
      p.second = count;
      pf.push_back(p);
    }
    i += 2;
  }
  if (b > 2) {
    p.first = b;
    p.second = 1;
    pf.push_back(p);
  }
  return pf;
}
long long int bs_t_i(string second) {
  long long int l = strtoull(second.c_str(), NULL, 2);
  return l;
}
string i_to_bs(long long int n) {
  string second = bitset<5>(n).to_string();
  return second;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int n;
  cin >> n;
  long long int a[n];
  long long int i = 0;
  while (i < n) {
    cin >> a[i];
    i++;
  }
  long long int sum1[n];
  long long int sumi = 0;
  i = 0;
  while (i < n) {
    sumi += a[i];
    sum1[i] = sumi;
    i++;
  }
  long long int q;
  cin >> q;
  while (q--) {
    long long int l, r;
    cin >> l >> r;
    l--;
    r--;
    cout << (sum1[r] - sum1[l] + a[l]) / 10 << endl;
  }
}

### Prompt

Please formulate a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long double pi = 3.14159265358979;
const long long int mod = 1e9 + 7;
long long int takemod(long long int a, long long int mod) {
  return ((a % mod) + mod) % mod;
}
long long int bs(long long int a[], long long int l, long long int r,
                 long long int x) {
  if (r >= l) {
    long long int mid = (l + r) / 2;
    if (a[mid] == x)
      return mid;
    else if (a[mid] > x)
      return bs(a, l, mid - 1, x);
    else
      return bs(a, mid + 1, r, x);
  }
  return -1;
}
long long int s_t_i(string second) {
  stringstream geek(second);
  long long int m;
  geek >> m;
  return m;
}
long long int power(long long int b, long long int p) {
  if (p == 0)
    return 1;
  else if (p % 2 == 0)
    return (power(b, p / 2) * (power(b, p / 2)));
  else
    return ((b) * (power(b, p - 1)));
}
bool isprime(long long int n) {
  if (n == 1) return false;
  if (n == 2 or n == 3) return true;
  long long int lim = sqrt(n);
  long long int i = 2;
  while (i <= lim) {
    if (n % i == 0) return false;
    i++;
  }
  return true;
}
vector<long long int> factors(long long int n) {
  vector<long long int> a;
  a.push_back(1);
  long long int lim = sqrt(n), i = 2;
  while (i <= lim) {
    if (n % i == 0) {
      a.push_back(i);
      if (i != n / i) a.push_back(n / i);
    }
    i++;
  }
  if (n != 1) a.push_back(n);
  sort(a.begin(), a.end());
  return a;
}
vector<pair<long long int, long long int> > pf_with_count(long long int b) {
  pair<long long int, long long int> p;
  vector<pair<long long int, long long int> > pf;
  long long int count = 0;
  while (b % 2 == 0) {
    b /= 2;
    count++;
  }
  if (count > 0) {
    p.first = 2;
    p.second = count;
    pf.push_back(p);
  }
  long long int i = 3;
  while (i <= sqrt(b)) {
    count = 0;
    while (b % i == 0) {
      b /= i;
      count++;
    }
    if (count > 0) {
      p.first = i;
      p.second = count;
      pf.push_back(p);
    }
    i += 2;
  }
  if (b > 2) {
    p.first = b;
    p.second = 1;
    pf.push_back(p);
  }
  return pf;
}
long long int bs_t_i(string second) {
  long long int l = strtoull(second.c_str(), NULL, 2);
  return l;
}
string i_to_bs(long long int n) {
  string second = bitset<5>(n).to_string();
  return second;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int n;
  cin >> n;
  long long int a[n];
  long long int i = 0;
  while (i < n) {
    cin >> a[i];
    i++;
  }
  long long int sum1[n];
  long long int sumi = 0;
  i = 0;
  while (i < n) {
    sumi += a[i];
    sum1[i] = sumi;
    i++;
  }
  long long int q;
  cin >> q;
  while (q--) {
    long long int l, r;
    cin >> l >> r;
    l--;
    r--;
    cout << (sum1[r] - sum1[l] + a[l]) / 10 << endl;
  }
}

```