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.lang.*;
import java.io.*;
import java.math.BigInteger;


public class CF {
	private static FS sc = new FS();
	
	private static class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());				               
	            } catch (IOException e) {}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	
	private static class extra {
		
		static int[] intArr(int size) {
			int[] a = new int[size];
			for(int i = 0; i < size; i++) a[i] = sc.nextInt();
			return a;
		}
		
		static long[] longArr(int size) {
			long[] a = new long[size];
			for(int i = 0; i < size; i++) a[i] = sc.nextLong();
			return a;
		}
		
		static long intSum(int[] a) {
			long sum = 0; 
			for(int i = 0; i < a.length; i++) {
				sum += a[i];
			}
			return sum;
		}
		
		static long longSum(long[] a) {
			long sum = 0; 
			for(int i = 0; i < a.length; i++) {
				sum += a[i];
			}
			return sum;
		}
		
		static LinkedList[] graphD(int vertices, int edges) {
			LinkedList[] temp = new LinkedList[vertices+1];
			for(int i = 0; i <= vertices; i++) temp[i] = new LinkedList<>();
			for(int i = 0; i < edges; i++) {
				int x = sc.nextInt();
				int y = sc.nextInt();
				temp[x].add(y);
			}
			return temp;
		}
		
		static LinkedList[] graphUD(int vertices, int edges) {
			LinkedList[] temp = new LinkedList[vertices+1];
			for(int i = 0; i <= vertices; i++) temp[i] = new LinkedList<>();
			for(int i = 0; i < edges; i++) {
				int x = sc.nextInt();
				int y = sc.nextInt();
				temp[x].add(y);
				temp[y].add(x);
			}
			return temp;
		}
		
		static void printG(LinkedList[] temp) {
			for(LinkedList<Integer> aa:temp) System.out.println(aa);
		}
		
	}
	
	static class sieve {
		static final int MAXN = (int) (Math.pow(10, 7)+5);
		
	    static int spf[] = new int[MAXN];
	    
	    static void sieve() {
	        spf[1] = 1;
	        for (int i=2; i<MAXN; i++) spf[i] = i;
	        for (int i=4; i<MAXN; i+=2) spf[i] = 2;
	        for (int i=3; i*i<MAXN; i++) if (spf[i] == i) for (int j=i*i; j<MAXN; j+=i) if (spf[j]==j) spf[j] = i;
	    }
	      
	    static long get(int x) {
	    	ArrayList<Integer> ret = new ArrayList<>();
	        while (x != 1) {
	            ret.add(spf[x]);
	            x = x / spf[x];
	        }
	        HashMap<Integer, Integer> temp = new HashMap<>();
	        for(int aa:ret) temp.put(aa, (temp.getOrDefault(aa, 0)+1)%2);
	        long pro = 1;
	        for(int aa:temp.keySet()) pro *= Math.pow(aa, temp.get(aa));
	        return pro;
	    }
	    
	}
	
	static LinkedList[] temp;
	static int mod = (int)Math.pow(10, 9) + 7;
	
	public static void main(String[] args) {	
//		int t = sc.nextInt();
		int t = 1;
		StringBuilder ret = new StringBuilder();
		while(t-- > 0) {
			int n = sc.nextInt();
			int[] a = extra.intArr(n);
			int q = sc.nextInt();
			long[] sum = new long[n+1];
			for(int i = 1; i <= n; i++) {
				sum[i] = sum[i-1]+a[i-1];
			}
			for(int i = 0; i < q; i++) {
				int l = sc.nextInt();
				int r = sc.nextInt();
				long s = sum[r] - sum[l-1];
				long quo = s/10;
				ret.append(quo + "\n");
			}
		}	
		System.out.println(ret);
	}
	
}

