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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.PrintStream;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Wolfgang Beyer
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        long MOD = 1000000007;

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            long[] prec = new long[100001];
            long current = 1;
            prec[0] = 1L;
            for (int i = 1; i <= 100000; i++) {
                current *= 2;
                current %= MOD;
                prec[i] = current;
            }

            int n = in.nextInt();
            int q = in.nextInt();
            String str = in.next();
            int[] s = new int[n + 1];
            for (int i = 1; i <= n; i++) {
                if (str.charAt(i - 1) == '1') {
                    s[i] = s[i - 1] + 1;
                } else {
                    s[i] = s[i - 1];
                }
            }
            for (int i = 0; i < q; i++) {
                int left = in.nextInt();
                int right = in.nextInt();
                int len = right - left + 1;
                int ones = s[right] - s[left - 1];
                int zeroes = len - ones;
                //long result = twoPow(len) - twoPow(zeroes);
                long result = prec[len] - prec[zeroes];
                result = (result + MOD) % MOD;
                out.println(result);
            }
        }

    }

    static class InputReader {
        private static BufferedReader in;
        private static StringTokenizer tok;

        public InputReader(InputStream in) {
            this.in = new BufferedReader(new InputStreamReader(in));
        }

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

        public String next() {
            try {
                while (tok == null || !tok.hasMoreTokens()) {
                    tok = new StringTokenizer(in.readLine());
                    //tok = new StringTokenizer(in.readLine(), ", \t\n\r\f"); //adds commas as delimeter
                }
            } catch (IOException ex) {
                System.err.println("An IOException was caught :" + ex.getMessage());
            }
            return tok.nextToken();
        }

    }
}