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.*;
import static java.lang.System.*;
import static java.lang.Math.*;
public class MainClass{
    public static void task(InputReader in){
        //---------------------------------Solve here---------------------------------------------------------------------------------
        int n=in.nextInt();
        int s[]=new int[n];
        int a[]=new int[n];
        int p=0;
        for(int i=0;i<n;i++)
        {
            s[i]=in.nextInt();
            p+=s[i];
            a[i]=p;
        }
        int q=in.nextInt();
        while(q-->0)
        {
            int l=in.nextInt();
            int r=in.nextInt();
            int i1=l-1;
            int i2=r-1;
            int value;
            if(i1-1<0)
            value=a[i2];
            else
            value=a[i2]-a[i1-1];
            int x=value/10;
            out.println(x);
        }
        //---------------------------------------------------------------------------------------------------------------------------- 
    }
    public static void main(String[] args)throws IOException {
        try{
            InputStream inputStream = System.in;
            InputReader in = new InputReader(inputStream);
            task(in);
            out.close();
        }
        catch(NumberFormatException e){
            System.out.println(e);
        }
    }
    public static class InputReader {
        /*
        String-nextString()
        int   -nextInt()
        double-nextDouble()
        long  -nextLong()
        char  -nextChar()
        line  -nextLine()
        */
		private InputStream stream;
		private byte[] buf = new byte[8192];
		private int curChar;
		private int numChars;
		private SpaceCharFilter filter;
		public InputReader(InputStream stream) {
			this.stream = stream;
		}
		public int read() {
			if (numChars == -1) {
				throw new InputMismatchException();
			}
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0) {
					return -1;
				}
			}
			return buf[curChar++];
		}
		public int nextInt() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
		public String nextString() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}
		public double nextDouble() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			double res = 0;
			while (!isSpaceChar(c) && c != '.') {
				if (c == 'e' || c == 'E') {
					return res * Math.pow(10, nextInt());
				}
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			}
			if (c == '.') {
				c = read();
				double m = 1;
				while (!isSpaceChar(c)) {
					if (c == 'e' || c == 'E') {
						return res * Math.pow(10, nextInt());
					}
					if (c < '0' || c > '9') {
						throw new InputMismatchException();
					}
					m /= 10;
					res += (c - '0') * m;
					c = read();
				}
			}
			return res * sgn;
		}
		public long nextLong() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
		public String nextLine() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isEndOfLine(c));
            return res.toString();
        }
        public char nextChar(){
            return nextString().charAt(0);
        }
		public boolean isSpaceChar(int c) {
			if (filter != null) {
				return filter.isSpaceChar(c);
			}
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
		public String next() {
			return nextString();
		}
		public interface SpaceCharFilter {
			public boolean isSpaceChar(int ch);
		}
		private boolean isEndOfLine(int c) {
            return c == '\n' || c == '\r' || c == -1;
        }
	}
}