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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;

import java.lang.management.MemoryType;
import java.math.BigInteger;
import java.util.*;
import java.util.function.Function;


public class Main {
    static final int INF = Integer.MAX_VALUE;

    static int mergeSort(int[] a,int [] c, int begin, int end)
    {
        int inversion=0;
        if(begin < end)
        {
            inversion=0;
            int mid = (begin + end) >>1;
            inversion+= mergeSort(a,c, begin, mid);
            inversion+=mergeSort(a, c,mid + 1, end);
            inversion+=merge(a,c, begin, mid, end);
        }
        return inversion;
    }


    static int  merge(int[] a,int[]c, int b, int mid, int e)
    {
        int n1 = mid - b + 1;
        int n2 = e - mid;
        int[] L = new int[n1+1], R = new int[n2+1];
        int[] L1 = new int[n1+1], R1 = new int[n2+1];


        //inversion
        int inversion=0;

        for(int i = 0; i < n1; i++) {
            L[i] = a[b + i];
            L1[i] = c[b + i];
          //  L2[i] = w[b + i];
        }
        for(int i = 0; i < n2; i++)  {
            R[i] = a[mid + 1 + i];
            R1[i] = c[mid + 1 + i];
            //R2[i] = w[mid + 1 + i];
        }
        L[n1] = R[n2] = INF;

        for(int k = b, i = 0, j = 0; k <= e; k++)
            if(L[i] <= R[j]){
                a[k] = L[i];
                c[k] = L1[i];
                ++i;
              //  w[k]=L2[i++];
            }

            else
            {
                a[k] = R[j];
                c[k] = R1[j];
               //w[k] = R2[j++];
                ++j;

                inversion=inversion+(n1-i);

            }
        return inversion;

    }

    static int []c;

    static int[] v;
    static long gcd(long a, long b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
    public class TreeNode {
        int val;
      TreeNode left;
      TreeNode right;
      TreeNode(int x) { val = x; }

  }
  int ans;
  void  dfs(TreeNode node,int L,int R){
        if (node !=null&&node.val>=L&&node.val<=R){
            ans+=node.val;
            if (node.val>L&&node.left!=null)dfs(node.left,L,R);
            if (node.val<R&&node.right!=null)dfs(node.right,L,R);
        }
  }

  static int bs(int i,int target,int []a,int l,int r ){
      int ans=i+a[l];
      while (r>=l){
          int mid=(r+l)/2;
          if (i+a[mid]>=target){
              r=mid-1;
          }else {
              ans=mid;
              l=mid+1;
          }
      }
      return ans;

  }
    static int bs2(int i,int target,int []a,int l,int r ){
        int ans=i+a[l];
        while (r>=l){
            int mid=(r+l)/2;
            if (i+a[mid]>=target){
                ans=mid;
                r=mid-1;
            }else {
                l=mid+1;
            }
        }
        return ans;

    }




     int rangeSumBST(TreeNode root, int L, int R) {
        if (root!=null)dfs(root,L,R);
        return ans;
    }



    public static void main (String[]args) throws IOException {
        Scanner in = new Scanner(System.in);
        try(PrintWriter or = new PrintWriter(System.out)){

            int n = in.nextInt();
            int[] a= new int[n];
            for(int i=0;i<n;++i){
                a[i]=in.nextInt();
                if (i!=0)a[i]+=a[i-1];
            }

            int m = in.nextInt();
            while (m-->0){
                int l=in.nextInt()-1;
                int r=in.nextInt()-1;
               if (l==0)or.println(a[r]/10);
               else or.println((a[r]-a[l-1])/10);
            }


        }



    }
    static int getlen(int r,int l,int a){
        return (r-l+1+1)/(a+1);
    }


    static class Scanner {

        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
        }

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++) {
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec) {
                        f *= 10;
                    }
                }
            }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public boolean ready() throws IOException {
            return br.ready();
        }

    }
}

 class Pair implements Comparable<Pair> {
    int first, second;

    public Pair(int first, int second) {
        this.first = first;
        this.second = second;
    }

    @Override
    public int compareTo(Pair o) {
        return first - o.first;
    }


}

class Tempo {

    int num;
   int index;

    public Tempo(int num, int index) {
        this.num = num;
        this.index = index;
    }
}