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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.StringTokenizer;

public class Round572C {
	
	public static class Pair{
		int first;
		int second;
	
		public Pair(int first,int second) {
			this.first=first;
			this.second=second;
		}
		
//		@Override
//		public boolean equals(Object obj) {
//			Pair temp=(Pair)obj;
//			return (temp.first==this.first && temp.second==this.second);
//		}
//		
//		@Override
//		public int hashCode() {
//			return (this.first * 0x1f1f1f1f) ^ this.second;
//		}
	}

	public static HashMap<Long,Pair> dp=new HashMap<Long,Pair>();
	
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		out=new PrintWriter (new BufferedOutputStream(System.out));
		FastReader s=new FastReader();
		int n=s.nextInt();
		int[] arr=new int[n];
		for(int i=0;i<n;i++)arr[i]=s.nextInt();
		for(int i=0;i<=17;i++) {
			if(i==0) {
				for(int j=0;j<n;j++) {
					dp.put(hashCode(j,j), new Pair(0,arr[j]));// first is candies and second is remaining
				}
			}else if(n>=(1<<i)) {
				int x=(1<<i);
				for(int j=0;j+x-1<n;j++) {
					int start=j;
					int end=j+x-1;
					int mid=(start+end)>>1;
					Pair left=dp.get(hashCode(start,mid));
					Pair right=dp.get(hashCode(mid+1,end));
					dp.put(hashCode(start,end), new Pair(left.first+right.first+(left.second+right.second)/10,(left.second+right.second)%10));
				}
			}
		}
		int q=s.nextInt();
		while(q-->0) {
			int l=s.nextInt();
			int r=s.nextInt();
			out.println(dp.get(hashCode(l-1,r-1)).first);
		}
		out.close();
	}

	public static long hashCode(long first,long second) {
		return (first*(long)1e5)+second;
	}
	
	public static PrintWriter out;
	
	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		//it reads the data about the specified point and divide the data about it ,it is quite fast
		//than using direct 

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
			//InputStream reads the data and decodes in character stream
			//It acts as bridge between byte stream and character stream
		}

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

		int nextInt() {
			return Integer.parseInt(next());//converts string to integer
		}

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

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

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