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 {

    public static void main(String[] args) throws Exception {

        Scanner sc = new Scanner();
        PrintWriter out = new PrintWriter(System.out);

        int MAX = (int)1e5 + 10, MOD = (int)1e9 + 7;
        int sum_powers[] = new int[MAX];
        int powers[] = new int[MAX];

        powers[0] = 1; sum_powers[0] = 1;
        for(int i = 1; i < MAX; i++){
            powers[i] = (int)(1L * powers[i - 1] * 2 % MOD);
            sum_powers[i] = sum_powers[i - 1] + powers[i];
            if (sum_powers[i] >= MOD) sum_powers[i] -= MOD;
        }


        int n = sc.nextInt(), q = sc.nextInt();
        int[] arr = new int[n], ones = new int[n];
        char[] x = sc.next().toCharArray();
        for(int i = 0; i < n; i++){
            arr[i] = x[i] - '0';
            ones[i] = arr[i];
            if(i > 0)ones[i] += ones[i - 1];
        }

        while (q-- > 0){
            int l = sc.nextInt() - 1, r = sc.nextInt() - 1;
            int ones_cnt = sum(ones, l, r), zeros_cnt = r - l + 1 - ones_cnt;
            if(ones_cnt == 0){
                out.println(0);
                continue;
            }
            int res = sum_powers[ones_cnt - 1];
            int res2 = (int)(1L * sum_powers[ones_cnt - 1] * (zeros_cnt == 0 ? 0 : sum_powers[zeros_cnt - 1]) % MOD);
            out.println((res + res2) % MOD);
        }

        out.flush();
        out.close();
    }

    static int sum (int[] a, int l, int r){
        return a[r] - (l == 0 ? 0 : a[l - 1]);
    }











    static class Scanner {
        BufferedReader br;
        StringTokenizer st;

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

        public Scanner(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(new File(s)));
        }

        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;
        }
    }
}