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 Main {
    static class Reader {
        BufferedReader in;

        Reader() throws IOException {
            in = new BufferedReader(new InputStreamReader(System.in)); // initial size buffer C = 32768 ??? /// default C = 8192
        }

        Reader(String name) throws IOException {
            in = new BufferedReader(new FileReader(name));
        }

        StringTokenizer tokin = new StringTokenizer("");

        String next() throws IOException {
            if (!tokin.hasMoreTokens()) {
                tokin = new StringTokenizer(in.readLine());
            }
            return tokin.nextToken();
        }

        int nextInt() throws IOException {
            return Integer.parseInt(next());
        }
    }

    static class Writer {
        PrintWriter cout;

        Writer() throws IOException {
            cout = new PrintWriter(System.out);
        }

        Writer(String name) throws IOException {
            cout = new PrintWriter(new FileWriter(name));
        }

        StringBuilder out = new StringBuilder();

        void print(Object a) {
            out.append(a);
        }

        void close() {
            cout.print(out.toString());
            cout.close();
        }
    }

    static long mod = Long.parseLong("1000000007");

    static long pow(long a, long b) {
        if (b == 0)
            return 1;
        if (b % 2 == 1) {
            return (a * pow(a, b - 1)) % mod;
        } else {
            long c = pow(a, b / 2);
            return (c * c) % mod;
        }
    }

    public static void main(String args[]) throws IOException {
        Reader in = new Reader();
        Writer out = new Writer();
        int n = in.nextInt();
        int q = in.nextInt();
        int col1[] = new int[n];
        String s = in.next();
        if (s.charAt(0) == '1')
            col1[0] = 1;
        for (int i = 1; i < n; i++) {
            col1[i] = col1[i - 1];
            if (s.charAt(i) == '1')
                col1[i]++;
        }
        for (int i = 0; i < q; i++) {
            int l = in.nextInt() - 1;
            int r = in.nextInt() - 1;
            long col = col1[r];
            long size = r - l + 1;
            if (l > 0)
                col -= col1[l - 1];
            long ans1 = pow(2, col) - 1;
            if (ans1 < 0)
                ans1 += mod;
            long p1 = pow(2, size - col) - 1;
            if (p1 < 0)
                p1 += mod;
            long ans = (ans1 + p1 * ans1) % mod;
            out.print(ans + "\n");
        }
        out.close();
    }
}