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.*;
import java.util.*;

public class C {

	void run() {
		int n = in.nextInt();
		int s[] = new int[n];
		int i,j;
		for(i=0;i<n;i++)
			s[i] = in.nextInt();

		//st[i][j] => (number, candy) of [i...i+2^j - 1]
		int maxJ = 0;
		while((1<<maxJ) <= n) maxJ++;
		int[][][] st = new int[n][maxJ][2];

		for(i=n-1;i>=0;i--){
			st[i][0][0] = s[i];
			st[i][0][1] = 0;
			for(j=1;j<maxJ;j++){
				int r = i + (1<<j) - 1;
				int jump = 1<<(j-1);
				if(r >= n) break;

				int n1 = st[i][j-1][0];
				int c1 = st[i][j-1][1];
				int n2 = st[i+jump][j-1][0];
				int c2 = st[i+jump][j-1][1];

				st[i][j][0] = (n1+n2)%10;
				st[i][j][1] = c1+c2+(n1+n2 >= 10 ? 1 : 0);
			}
		}

		// querys
		int q = in.nextInt();
		for(i=0;i<q;i++){
			int l = in.nextInt() - 1;
			int r = in.nextInt() - 1;
			// r = l + 1<<j - 1
			j = 0;
			while(l + (1<<j) - 1 != r)
				j++;
			
			int ans = st[l][j][1];
			out.println(ans);
		}
	}
	
	static MyScanner in;
	static PrintWriter out;

	public static void main(String[] args) throws IOException {
		boolean stdStream = true;
		if(stdStream){
			in = new MyScanner();
			out = new PrintWriter(System.out);
		}else{
			String fileName = C.class.getSimpleName();
			int n_test = 1;			
			String inputFileName = fileName + String.format(".%02d", n_test) + ".inp";
			String outputFileName = fileName + String.format(".%02d", n_test) + ".out";
			in = new MyScanner(new BufferedReader(new FileReader(inputFileName)));
			out = new PrintWriter(outputFileName);
		}
			
		new C().run();

		in.close();
		out.close();
	}

	static class MyScanner {
		BufferedReader br;
		StringTokenizer st;

        MyScanner(){
            this.br = new BufferedReader(new InputStreamReader(System.in));
        }

		MyScanner(BufferedReader br) {
			this.br = br;
        }

        void close() throws IOException {
            br.close();
        }
        
		void findToken() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
		}
		
		String next() {
			findToken();
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
}