JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.

First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.

For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.

However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.

All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).

The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.

Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.

Output

Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.

Examples

Input

4 2
1011
1 4
3 4


Output

14
3


Input

3 2
111
1 2
3 3


Output

3
1

Note

In the first example: 

  * For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2. 
  * For query 2: Both 3, 4 and 4, 3 ordering give the same answer. 



In the second example, any order of eating parts leads to the same answer.

 import java.io.*;

import java.util.*;
import java.util.Collections;
import java.util.Arrays;


public class Codechef {
       
  public static void main(String[] args) throws IOException {

    Scanner sc = new Scanner(System.in);
       long powers[]=new long[100001];
              powers[0]=1;
        for(int i=1;i<100001;i++){
            powers[i]=(2*powers[i-1])%1000000007;
        }
         
          int n=sc.nextInt();
          int q=sc.nextInt();
          String s=sc.next();

                       int[] a=new int[n+1];

             for(int h=1;h<=n;h++){
            if(s.charAt(h-1)=='1'){
                a[h]=a[h-1]+1;
            }else a[h]=a[h-1];

            //System.out.print("a["+(h)+"] = "+a[h]+" ");
        }
          for(int i=0;i<q;i++){
            int l=sc.nextInt();
            int r=sc.nextInt();
             int ones=a[r]-a[l-1];
             int zeros=r-l+1-ones;
            long onesum=powers[ones]-1;
        long zerosum=(onesum*(powers[zeros]))%1000000007;
        System.out.println(zerosum);

         // System.out.println("ones:"+ones+" zeros: "+zeros);


            // long sum=(long)Math.pow(2,k.length())-1;
            // long sum2=(long)Math.pow(2,count)-1;
           // long result=powers(2,r-l+1)-powers(2,zeros);
            
            //System.out.println(result%1000000007);
            

                            
                                
                                                       

                             
                        
          }

          





          }
  }








