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.
//make sure to make new file!
import java.io.*;
import java.util.*;

public class C520{

   public static long MOD = 1000000007L;
   
   public static void main(String[] args)throws IOException{
      BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter out = new PrintWriter(System.out);
      
      
      long[] pow2 = new long[100005];
      pow2[0] = 1L;
      for(int k = 1; k < 100005; k++){
         pow2[k] = (2L*pow2[k-1] + MOD)%MOD;
      }
      
      
      StringTokenizer st = new StringTokenizer(f.readLine());
      
      int n = Integer.parseInt(st.nextToken());
      int m = Integer.parseInt(st.nextToken());
      
      char[] sarray = f.readLine().toCharArray();
      int[] array = new int[n];
      for(int k = 0; k < n; k++){
         if(sarray[k] == '1') array[k] = 1;
         else array[k] = 0;
      }
         
      int[] psum = new int[n+1];
         
      psum[0] = 0;
      for(int k = 1; k <= n; k++){
         psum[k] = psum[k-1] + array[k-1];
      }
         
      for(int k = 0; k < m; k++){
         st = new StringTokenizer(f.readLine());
         int l = Integer.parseInt(st.nextToken());
         int r = Integer.parseInt(st.nextToken());
            
         int b = psum[r] - psum[l-1];
         int a = r-l+1-b;
            
         long answer = ((pow2[b]-1)*pow2[a] + MOD)%MOD;
         out.println(answer);
      }
      
      
      
      
      
      out.close();
   }
   
      
}