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:
//package Div2_520;

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

public class C {
    public static BufferedReader input;
    public static PrintWriter out;
    public StringTokenizer stoken = new StringTokenizer("");

    public static void main(String[] args) throws IOException {
        input = new BufferedReader(
                new InputStreamReader(System.in)
        );
        out = new PrintWriter(System.out);
        new C();
        out.close();
    }

    C() throws IOException {
        int n = nextInt();
        int q = nextInt();
        long mod = 1000_000_007;
        long s[] = new long[n + 1];
        s[0] = 1;
        for (int i = 1; i < n + 1; i++) {
            s[i] = s[i - 1] * 2;
            s[i] %= mod;
        }
        String c = nextString();
        int mas[] = new int[n];
        for (int i = 0; i < n; i++) {
            mas[i] = Integer.parseInt(String.valueOf(c.charAt(i)));
        }
        int pref[] = new int[n + 1];
        pref[1] = mas[0];
        for (int i = 1; i < n; i++) {
            pref[i + 1] += pref[i] + mas[i];
        }
        for (int i = 0; i < q; i++) {
            int l = nextInt();
            int r = nextInt();
            int d = r - l + 1;
            int one = pref[r] - pref[l - 1];
            int nu = d - one;
            long ans = s[one] - 1 + (s[one] - 1) * (s[nu] - 1);
            ans %= mod;
            out.println(ans);
        }
    }

    private String nextString() throws IOException {
        while (!stoken.hasMoreTokens()) {
            String st = input.readLine();
            stoken = new StringTokenizer(st);
        }
        return stoken.nextToken();
    }

    private Integer nextInt() throws IOException {
        return Integer.parseInt(nextString());
    }
}