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.
//package codeforces.practice.div2c;

import java.io.*;
import java.util.*;

public class CF1189C {
    public static void main(String[] args) {
//        try {
//            FastScanner in = new FastScanner(new FileInputStream("src/input.in"));
//            PrintWriter out = new PrintWriter(new FileOutputStream("src/output.out"));

        FastScanner in = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

            solve(1, in, out);
//        } catch (IOException e) {
//            e.printStackTrace();
//        }
    }

    private static void solve(int q, FastScanner in, PrintWriter out) {
        for (int qq = 0; qq < q; qq++) {
            int n = in.nextInt();
            Integer[] a = in.nextIntArray(n);
            int[] ps = new int[n];
            ps[0] = a[0];
            for(int i = 1; i < n; i++) {
                ps[i] = ps[i - 1] + a[i];
            }
            int t = in.nextInt();
            for(int i = 0; i < t; i++) {
                int l = in.nextInt() - 1, r = in.nextInt() - 1;
                out.println((ps[r] - (l > 0 ? ps[l - 1] : 0)) / 10);
            }
        }
        out.close();
    }

    private static class NumberTheory {
        private static long modularAdd(long a, long b, int mod) {
            long sum = a + b;
            if (sum >= mod) {
                sum -= mod;
            }
            return sum;
        }

        private static long modularSubtract(long a, long b, int mod) {
            long diff = a - b;
            if (diff < 0) {
                diff += mod;
            }
            return diff;
        }

        private static long fastModPow2(long x, int n, int mod) {
            if (n == 0) {
                return 1;
            }
            long coeff = 1, base = x;
            while (n > 1) {
                if (n % 2 != 0) {
                    coeff = (coeff * base % mod);
                }
                base = (base * base % mod);
                n = n / 2;
            }
            long res = coeff * base % mod;
            return res;
        }

        private static long gcd(long a, long b) {
            if (b == 0) {
                return a;
            }
            return gcd(b, a % b);
        }
    }

    private static class Graph {
        private static int UNVISITED = 0;
        private static int VISITING = -1;
        private static int VISITED = 1;

        private static Stack<Integer> topologicalSort(Set<Integer>[] g) {
            Stack<Integer> stack = new Stack<>();
            int[] state = new int[g.length];

            for (int node = 0; node < g.length; node++) {
                if (!topoSortHelper(g, stack, state, node)) {
                    return null;
                }
            }
            return stack;
        }

        private static boolean topoSortHelper(Set<Integer>[] g, Stack<Integer> stack, int[] state, int currNode) {
            if (state[currNode] == VISITED) {
                return true;
            } else if (state[currNode] == VISITING) {
                return false;
            }
            state[currNode] = VISITING;
            for (int neighbor : g[currNode]) {
                if (!topoSortHelper(g, stack, state, neighbor)) {
                    return false;
                }
            }
            state[currNode] = VISITED;
            stack.push(currNode);
            return true;
        }
    }

    private static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream stream) {
            try {
                br = new BufferedReader(new InputStreamReader(stream));
            } catch (Exception e) {
                e.printStackTrace();
            }
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        Integer[] nextIntArray(int n) {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextInt();
            }
            return a;
        }

        Long[] nextLongArray(int n) {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextLong();
            }
            return a;
        }
    }
}


