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.stream.Stream;

/**
 * @author madi.sagimbekov
 */
public class C1062C {

    private static BufferedReader in;
    private static BufferedWriter out;

    public static void main(String[] args) throws IOException {
        open();

        int[] nq = readInts();
        int n = nq[0];
        int q = nq[1];

        char[] a = readString().toCharArray();

        int[] z = new int[n];
        int[] o = new int[n];

        for (int i = 0; i < n; i++) {
            if (i == 0) {
                if (a[i] == '0') {
                    z[i]++;
                } else {
                    o[i]++;
                }
            } else {
                if (a[i] == '0') {
                    z[i] = z[i - 1] + 1;
                    o[i] = o[i - 1];
                } else {
                    o[i] = o[i - 1] + 1;
                    z[i] = z[i - 1];
                }
            }
        }

        long mod = 1000000007;

        for (int i = 0; i < q; i++) {
            int[] lr = readInts();
            int l = lr[0];
            int r = lr[1];

            int zeros = z[r - 1] - (l == 1 ? 0 : z[l - 2]);
            int ones = o[r - 1] - (l == 1 ? 0 : o[l - 2]);

            long ans1 = pow(ones, mod) - 1;
            long ans2 = pow(zeros, mod) - 1;
            long ans = (ans1 % mod + (ans1 * ans2) % mod) % mod;

            out.write(ans + "\n");

        }

        close();
    }

    private static long pow(int k, long mod) {
        if (k == 0) {
            return 1;
        }
        if (k == 1) {
            return 2;
        }

        long res = pow(k / 2, mod);
        if (k % 2 == 0) {
            return (res * res) % mod;
        } else {
            return (res * res * 2) % mod;
        }
    }

    private static int[] readInts() throws IOException {
        return Stream.of(in.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
    }

    private static int readInt() throws IOException {
        return Integer.parseInt(in.readLine());
    }

    private static long[] readLongs() throws IOException {
        return Stream.of(in.readLine().split(" ")).mapToLong(Long::parseLong).toArray();
    }

    private static long readLong() throws IOException {
        return Long.parseLong(in.readLine());
    }

    private static double[] readDoubles() throws IOException {
        return Stream.of(in.readLine().split(" ")).mapToDouble(Double::parseDouble).toArray();
    }

    private static double readDouble() throws IOException {
        return Double.parseDouble(in.readLine());
    }

    private static String readString() throws IOException {
        return in.readLine();
    }

    private static void open() {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new BufferedWriter(new OutputStreamWriter((System.out)));
    }

    private static void close() throws IOException {
        out.flush();
        out.close();
        in.close();
    }

}