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.net.Inet4Address;
import java.util.*;
import java.io.*;

public class Did_I_Deserve_It{

    public static long power(long x, long y, long p)
    {
        // Initialize result
        long res = 1;

        // Update x if it is more
        // than or equal to p
        x = x % p;

        while (y > 0)
        {
            // If y is odd, multiply x
            // with result
            if((y & 1)==1)
                res = (res * x) % p;

            // y must be even now
            // y = y / 2
            y = y >> 1;
            x = (x * x) % p;
        }
        return res;
    }

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

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);

        StringTokenizer st1 = new StringTokenizer(br.readLine());

        int n = Integer.parseInt(st1.nextToken());
        int q = Integer.parseInt(st1.nextToken());

        String ss = br.readLine();

        point arr[] = new point[n];

        long zeros = 0l, ones = 0l;
        for(int i = 0 ; i < ss.length() ; i++)
        {
            if(ss.charAt(i) == '1')
                ones+=1l;
            else
                zeros+=1l;

            arr[i] = new point(zeros , ones);


        }

        while(q-- > 0)
        {
            st1 = new StringTokenizer(br.readLine());

            int l = Integer.parseInt(st1.nextToken())-1;
            int r = Integer.parseInt(st1.nextToken())-1;

            ones = arr[r].ones - arr[l].ones;
            zeros = arr[r].zeros - arr[l].zeros;

            if(ss.charAt(l) == '0')
                zeros+=1l;
            else
                ones+=1l;


            long sum = (long)(power(2,ones,1000000007)-1l)*(power(2,zeros,1000000007));

            out.println(sum % (1000000007));



        }


        out.flush();
        out.close();

    }

    static class point{

        long zeros , ones;

        public point(long z , long o)
        {
            zeros = z;
            ones = o;
        }
    }

}