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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main implements Runnable {
    public static void main (String args[]){
        new Thread(null, new Main(), "Main", 1 << 26).start();
//        new Main().run();
    }

    static final int BASE = (int)1e9 + 7;
    static int[] cul;
    static int[] f;

    public void run() {
        MyScanner sc = new MyScanner();

        int n = sc.nextInt();
        int q = sc.nextInt();
        String s = sc.nextLine();

        int[] count = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            count[i] = count[i - 1];
            if (s.charAt(i - 1) == '1') count[i]++;
        }

        init(n);

        for (int i = 0; i < q; i++) {
            int l = sc.nextInt();
            int r = sc.nextInt();

            int c1 = count[r] - count[l - 1];
            int c0 = r - l + 1 - c1;

            long huy = ((long)cul[c1 + c0] - cul[c1] - cul[c0]) % BASE;
            if (huy < 0) huy += BASE;
            System.out.println((cul[c1] + huy) % BASE);
        }
    }

    void init(int n) {
        n++;
        f = new int[n + 1];
        cul = new int[n + 1];

        f[1] = 1;
        for (int i = 2; i <= n; i++) {
            f[i] = (f[i - 1] * 2) % BASE;
        }

        for (int i = 1; i <= n; i++) {
            cul[i] = (cul[i - 1] + f[i]) % BASE;
        }
    }
}

class MyScanner {
    BufferedReader br;
    StringTokenizer st;

    public MyScanner() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

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

    long nextLong() {
        return Long.parseLong(next());
    }

    double nextDouble() {
        return Double.parseDouble(next());
    }

    String nextLine(){
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }

}