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.util.*;

public class BanhMi
{
    PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));

    long MOD = 1000000007;

    public void go()
    {
        int numParts = in.nextInt();
        int numQueries = in.nextInt();
        int[] oneCount = new int[numParts+1];
        int[] zeroCount = new int[numParts+1];
        char[] input = in.next().toCharArray();
        for (int i = 1; i <= numParts; i++)
        {
            oneCount[i] = oneCount[i-1] + (input[i-1] == '1' ? 1 : 0);
            zeroCount[i] = zeroCount[i-1] + (input[i-1] == '0' ? 1 : 0);
        }
        for (int i = 0; i < numQueries; i++)
        {
            int low = in.nextInt();
            int high = in.nextInt();
            int ones = oneCount[high] - oneCount[low-1];
            int zeros = zeroCount[high] - zeroCount[low-1];
            long ans = (pow(2, ones) + MOD-1) % MOD;
            // out.println("    " + ans);
            ans = (ans * pow(2, zeros)) % MOD;
            out.println(ans);
        }

        out.flush();
        in.close();
    }

    public long pow(long x, int n)
    {
        if (n == 0)
        {
            return 1;
        }
        long x2 = x * x % MOD;
        return pow(x2, n/2) * (n % 2 == 0 ? 1 : x) % MOD;
    }

    public static void main(String[] args)
    {
        new BanhMi().go();
    }
}