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 long[] degs;

    public static void main(String[] args) {

        try (BufferedReader reader = new BufferedReader(new InputStreamReader(System.in))) {

            String[] line = reader.readLine().split(" ");
            int n = Integer.parseInt(line[0]);
            int q = Integer.parseInt(line[1]);

            char[] ch = reader.readLine().toCharArray();
            int[] a = new int[n + 1];
            a[1] = ch[0] - '0';
            for(int i = 2; i <= n; i++) {
                a[i] = a[i - 1] + (ch[i - 1] - '0');
            }
            degs = new long[n + 1];
            deg(2, n + 1);

            for(int i = 0; i < q; i++) {
                line = reader.readLine().split(" ");
                int l = Integer.parseInt(line[0]);
                int r = Integer.parseInt(line[1]);
                long oneSum = degs[a[r] - a[l - 1]] - 1;
                long secSum = (degs[(r - l + 1) - (a[r] - a[l - 1])] - 1) * oneSum;
                //println(oneSum + " " + secSum);
                println((oneSum + secSum) % 1_000_000_007);
            }

        } catch (IOException e) {
        }
    }

    static long deg(int num, int deg) {
        long ans = 1;
        for(int i = 0; i < deg; i++) {
            degs[i] = ans;
            ans *= num;
            ans %= 1_000_000_007;
        }
        return ans;
    }

    static String reverse(String s) {
        String ans = "";
        char[] c = s.toCharArray();
        for(int i = s.length() - 1; i >= 0; i--) {
            ans += c[i];
        }
        return ans;
    }

    static long fac(int deg) {
        long ans = 1;
        for(int i = 2; i <= deg; i++) {
            ans *= i;
        }
        return ans;
    }

    static int r(int a, int b) {
        int ans = a / b;
        if(a % b != 0)
            ans++;
        return ans;
    }

    static void print(Object o) {
        System.out.print(o.toString());
    }

    static void println(Object o) {
        System.out.println(o.toString());
    }
}