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.lang.*;
import java.util.*;
import java.io.*;

public class CF1335{
    static long power(long x, long y, long p) 
    { 
        long res = 1;      
        x = x % p;  
  
       if (x == 0) return 0;   
  
        while (y > 0) 
        { 
            if((y & 1)==1) 
                res = (res * x) % p; 
            y = y >> 1;  
            x = (x * x) % p;  
        } 
        return res; 
    } 
    static long calculate(int ones,int zeros){
        if(ones==0){
            return 0;
        }
        long ans = 0;
        long mod = 1000000007;
        ans = power(2,ones,mod);
        ans--;
        long val = ans;
        long ans2 = power(2,zeros,mod);
        ans2--;
        ans2 = ((ans2%mod)*(val%mod))%mod;
        ans = ((ans%mod)+(ans2%mod))%mod;
        return ans;
    }
    public static void main(String[] args)throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = new Integer(st.nextToken());
        int q = new Integer(st.nextToken());
        StringBuffer sb = new StringBuffer();
        String a = br.readLine();
        int[] ones = new int[a.length()+1];
        int[] zeros = new int[a.length()+1];
        for(int i=0;i<a.length();i++){
            if(a.charAt(i)=='1'){
                ones[i+1]=ones[i]+1;
                zeros[i+1]=zeros[i];
            }
            else{
                zeros[i+1]=zeros[i]+1;
                ones[i+1]=ones[i];
            }
        }
        while(q-->0){
            StringTokenizer st1 = new StringTokenizer(br.readLine());
            int w = new Integer(st1.nextToken()),v = new Integer(st1.nextToken());
            sb.append(calculate(ones[v]-ones[w-1],zeros[v]-zeros[w-1])).append("\n");
        }
        System.out.println(sb);
    }
}