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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

/**
 * Created by Katushka on 31.01.2020.
 */
public class BahnMi {

    private static long fastPow2(int y) {
        if (y == 0) {
            return 1;
        }
        long p = fastPow2(y / 2) % 1000000007;
        p = p * p % 1000000007;
        if (y % 2 == 1) {
            p = p * 2 % 1000000007;
        }
        return p;
    }

    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        int n = in.nextInt();
        int q = in.nextInt();
        String s = in.nextString();
        int[] zeroes = new int[s.length()];
        int cur = 0;

        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '0') {
                cur += 1;

            }
            zeroes[i] = cur;
        }
        for (int i = 0; i < q; i++) {
            int l = in.nextInt();
            int r = in.nextInt();
            int n0;
            if (l == 1) {
                n0 = zeroes[r - 1];
            } else {
                n0 = zeroes[r - 1] - zeroes[l - 2];
            }
            long p1 = fastPow2(r - l + 1) % 1000000007;
            long p2 = fastPow2(n0) % 1000000007;
            if (p1 >= p2) {
                System.out.println(p1 - p2);
            } else{
                System.out.println(p1 + 1000000007 - p2);
            }
        }

    }

    private static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    final String str = reader.readLine();
                    tokenizer = new StringTokenizer(str);
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public String nextString() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }

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

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

        public char nextChar() {
            return next().charAt(0);
        }
    }
}