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

public class Candies {
    PrintWriter out;
    StringTokenizer st;
    BufferedReader br;
    final int imax = Integer.MAX_VALUE, imin = Integer.MIN_VALUE;
    final int mod = 1000000007;

    void solve() throws Exception {
        int t = 1;
//        t = ni();
        for (int ii = 0; ii < t; ii++) {
            int n= ni();
            int[] a= ni(n);

            for (int i = 0; i < n; i++) a[i]%= 10;
            int[] pre= new int[n];
            for (int i = 0; i < n; i++) pre[i]= (i!= 0? pre[i-1]: 0)+ a[i];
            
            int q= ni();
            for (int i = 0; i < q; i++) {
                int s= ni()-1, l= ni()-1;
                out.println((pre[l]-(s!= 0?pre[s-1]: 0))/10);
            }
        }
    }

    public static void main(String[] args) throws Exception {
        new Candies().run();
    }

    void run() throws Exception {
        if (System.getProperty("ONLINE_JUDGE") == null) {
            File file = new File("C:\\college\\CodeForces\\inputf.txt");
            br = new BufferedReader(new FileReader(file));
            out = new PrintWriter("C:\\college\\CodeForces\\outputf.txt");
        } else {
            out = new PrintWriter(System.out);
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        long ss = System.currentTimeMillis();
        st = new StringTokenizer("");
        while (true) {
            solve();
            String s = br.readLine();
            if (s == null) break;
            else st = new StringTokenizer(s);
        }
        //out.println(System.currentTimeMillis()-ss+"ms");
        out.flush();
    }

    void read() throws Exception {
        st = new StringTokenizer(br.readLine());
    }

    int ni() throws Exception {
        if (!st.hasMoreTokens()) read();
        return Integer.parseInt(st.nextToken());
    }

    char nc() throws Exception {
        if (!st.hasMoreTokens()) read();
        return st.nextToken().charAt(0);
    }

    long nl() throws Exception {
        if (!st.hasMoreTokens()) read();
        return Long.parseLong(st.nextToken());
    }

    int[] ni(int n) throws Exception {
        int[] ret = new int[n];
        for (int i = 0; i < n; i++) ret[i] = ni();
        return ret;
    }

    long[] nl(int n) throws Exception {
        long[] ret = new long[n];
        for (int i = 0; i < n; i++) ret[i] = nl();
        return ret;
    }

    double nd() throws Exception {
        if (!st.hasMoreTokens()) read();
        return Double.parseDouble(st.nextToken());
    }

    String ns() throws Exception {
        String s = br.readLine();
        return s.length() == 0 ? br.readLine() : s;
    }

    void print(int[] arr) {
        for (int i : arr) out.print(i + " ");
        out.println();
    }

    void print(long[] arr) {
        for (long i : arr) out.print(i + " ");
        out.println();
    }

    void print(int[][] arr) {
        for (int[] i : arr) {
            for (int j : i) out.print(j + " ");
            out.println();
        }
    }

    void print(long[][] arr) {
        for (long[] i : arr) {
            for (long j : i) out.print(j + " ");
            out.println();
        }
    }

    long add(long a, long b) {
        if (a + b >= mod) return (a + b) - mod;
        else return a + b >= 0 ? a + b : a + b + mod;
    }

    long mul(long a, long b) {
        return (a * b) % mod;
    }

    void print(boolean b) {
        if (b) out.println("YES");
        else out.println("NO");
    }

    long binExp(long base, long power) {
        long res = 1l;
        while (power != 0) {
            if ((power & 1) == 1) res = mul(res, base);
            base = mul(base, base);
            power >>= 1;
        }
        return res;
    }

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

    // strictly smaller on left
    void stack_l(int[] arr, int[] left) {
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < arr.length; i++) {
            while (!stack.isEmpty() && arr[stack.peek()] >= arr[i]) stack.pop();
            if (stack.isEmpty()) left[i] = -1;
            else left[i] = stack.peek();
            stack.push(i);
        }
    }

    // strictly smaller on right
    void stack_r(int[] arr, int[] right) {
        Stack<Integer> stack = new Stack<>();
        for (int i = arr.length - 1; i >= 0; i--) {
            while (!stack.isEmpty() && arr[stack.peek()] >= arr[i]) stack.pop();
            if (stack.isEmpty()) right[i] = arr.length;
            else right[i] = stack.peek();
            stack.push(i);
        }
    }

    private void sort(int[] arr) {
        List<Integer> list = new ArrayList<>();
        for (int i : arr) list.add(i);
        Collections.sort(list);
        for (int i = 0; i < arr.length; i++) arr[i] = list.get(i);
    }
}