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 java.math.*;
import java.lang.*;
 
import static java.lang.Math.*;
 
public class Solution implements Runnable 
{
	static class InputReader 
	{
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;
		private SpaceCharFilter filter;
		private BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
 
		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 String nextLine()
		{
			String str = "";
			try
			{
				str = br.readLine();
			}
			catch (IOException e)
			{
				e.printStackTrace();
			}	
			return str;
		}
		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 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 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 String readString() 
		{
			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 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 readString();
		}
        
		public interface SpaceCharFilter
		{
			public boolean isSpaceChar(int ch);
		}
		
		public int[] readint(int st,int n){
		    int[] arr=new int[n+st];
		    for(int i=st;i<n+st;i++) arr[i]=nextInt();
		    return arr;
		}
		
		public long[] readlong(int st,int n){
		    long[] arr=new long[n+st];
		    for(int i=st;i<n+st;i++) arr[i]=nextLong();
		    return arr;
		}
	}
	public static void main(String args[]) throws Exception 
	{
		new Thread(null, new Solution(),"Main",1<<27).start();
	}	
	
	public static long gcd(long a, long b) 
	{ 
		if (a == 0) 
			return b; 
		return gcd(b % a, a); 
	} 
  
	public static long findGCD(long arr[], int n) 
	{ 
		long result = arr[0]; 
		for (int i = 1; i < n; i++) 
			result = gcd(arr[i], result); 
		return result; 
	}	 
	
	static void sortbycolomn(long arr[][], int col)
    { 
         
        Arrays.sort(arr, new Comparator<long[]>() { 
            
          @Override              
           
          public int compare(final long[] entry1,  
                             final long[] entry2) { 
  
            
            if (entry1[col] > entry2[col]) 
                return 1; 
            else
                return -1; 
          } 
        });   
    } 
	
	public void run()
	{
		InputReader in = new InputReader(System.in);
		PrintWriter w = new PrintWriter(System.out);
		
		int n = in.nextInt();
		int[] arr=new int[n];
		int x=0;
		for(int i=0;i<n;i++){
		    arr[i] = x + in.nextInt();
		    x = arr[i];
		}
		int q = in.nextInt();
		for(int i=0;i<q;i++){
		    int l = in.nextInt();
		    int r = in.nextInt();
		    if(l!=1) l = arr[l-2];
		    else l=0;
		    r = arr[r-1];
		    int ans = (r-l)/10;
		    w.println(ans);
		}
		w.flush();
		w.close();
	}
}