Problem: 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.

Solution (in JAVA):

Solution:
import java.io.*;
import java.lang.reflect.Array;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;

public class Solution {
    
    private static final Scanner sc = new Scanner(System.in);

    public static void main(String[] args) throws IOException {
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int q = Integer.parseInt(st.nextToken());
        String s = br.readLine();   
        long mod = 1000000007;
        long [] p2s = new long[100002];
        p2s[0]=1;
        for (int i=1; i<p2s.length; ++i){
            p2s[i]=p2s[i-1]*2;
            p2s[i]%=mod;
        }
        int[] arr = new int[n];
        int[] acc = new int[n+1];
        for (int i=0; i<n; ++i){
            arr[i] = s.charAt(i)-'0';
            acc[i+1] = acc[i] + arr[i];    
        }
        for (int qq=0; qq<q; ++qq){
            st = new StringTokenizer(br.readLine());
            int l = Integer.parseInt(st.nextToken());
            int r = Integer.parseInt(st.nextToken());
            int d = r-l+1;
            int c = acc[r]-acc[l-1];
            if (c == 0){
                bw.write(""+0);
                bw.newLine();
            } else {
                long cc = ((p2s[c]+mod)-1)%mod;
                long cc2 = ((p2s[d-c]+mod)-1)%mod;
                cc2*= cc;
                cc2%=mod;
                cc+=cc2;
                cc%=mod;
                
                
                bw.write(""+cc);
                bw.newLine();
            }
        }
        
        sc.close();
        br.close();
        bw.close();
    }
    
}