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.io.*;
import java.util.Arrays;
import java.util.StringTokenizer;

import static java.lang.Math.*;

public class Main {
    FastScanner in;
    PrintWriter out;

    void run() {
        in = new FastScanner();
        out = new PrintWriter(System.out);
        problem();
        out.close();
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

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

        public FastScanner(String s) {
            try {
                br = new BufferedReader(new FileReader(s));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }
        }

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

        public int nextInt() {
            return Integer.parseInt(nextToken());
        }

        public long nextLong() {
            return Long.parseLong(nextToken());
        }

        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }

    public static void main(String[] args) {
        new Main().run();
    }

    void build (int a[], int v, int tl, int tr) {
        if (tl == tr)
            t[v] = a[tl];
        else {
            int tm = (tl + tr) / 2;
            build(a, v * 2, tl, tm);
            build(a, v * 2 + 1, tm + 1, tr);
            t[v] = t[v * 2] + t[v * 2 + 1];
        }
    }

    int sum (int v, int tl, int tr, int l, int r) {
        if (l > r)
            return 0;
        if (l == tl && r == tr)
            return t[v];
        int tm = (tl + tr) / 2;
        return sum (v*2, tl, tm, l, min(r,tm))
                + sum (v*2+1, tm+1, tr, max(l,tm+1), r);
    }

    public static long modPower(int x, int y)
    {
        if(y==0) return 1;
        long z = modPower(x, y/2);
        if ((y % 2) == 0)
            return (z*z)%N;
        else
            return (x*z*z)%N;
    }

    int [] t;
    static int N = (int)pow(10, 9) + 7;


    void problem() {
        int n = in.nextInt();
        int q = in.nextInt();
        String s = in.nextToken();

        int a[] = new int[n];

        for (int i = 0; i < n; i++) {
            a[i] = Character.getNumericValue(s.charAt(i));
        }

        t = new int [4*n];

        build(a, 1, 0, n - 1);

        for (int asd = 0; asd < q; asd++) {
            int l = in.nextInt();
            int r = in.nextInt();

            int k = sum(1, 0, n - 1, l - 1, r - 1);
            int z = r - l + 1 - k;

            long a1 = modPower(2, k) - 1;
            long sum0 = ((modPower(2, z) - 1) * a1)  % N;

            out.println((a1 + sum0) % N);
        }
    }
}
