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.

import java.util.Scanner;

public class C {

    static long m(int n) {
        if (n == 0) {
            return 1;
        }
        if (n % 2 == 1) {
            long t = m(n / 2);
            return 2 * t * t % 1000000007;
        } else {
            long t = m(n / 2);
            return t * t % 1000000007;
        }
    }

    static long work(int sum) {
        return m(sum) - 1;
    }

    public static void main(String[] aaaa) {
        Scanner scanner = new Scanner(System.in);

        int n = scanner.nextInt();
        int q = scanner.nextInt();

        byte[] a = scanner.next().getBytes();

        //int[] data = new int[100005];
        int[] tree = new int[100005];


        for (int i = 0; i < a.length; i++) {
            if (a[i] == '1') {
                //data[i + 1] = i;
                int j = i + 1;

                while (j <= n) {
                    tree[j]++;
                    j += (j & (-j));
                }
            }
        }

        StringBuilder stringBuilder = new StringBuilder();
        for (int i = 0; i < q; i++) {
            int l = scanner.nextInt();
            int r = scanner.nextInt();

            //for (int iii = 0; iii < 100000; iii++) {
                int j = r;
                int sum = 0;
                while (j > 0) {
                    sum += tree[j];
                    j -= (j & (-j));
                }

                j = l - 1;
                while (j > 0) {
                    sum -= tree[j];
                    j -= (j & (-j));
                }

                //System.out.println(sum);

                long ans = 0;
                int len = (r - l + 1);
                //ans += (sum * (sum + 1)) / 2;
                long tmp = work(sum);

                ans += tmp;
                ans += tmp * work(len - sum);
                //System.out.println(ans % 1000000007);
            stringBuilder.append(ans % 1000000007);
            stringBuilder.append("\n");
            //}


        }
        System.out.println(stringBuilder.toString());





    }
}
