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.*;
import java.math.*;

public class A implements Runnable {

    public void run() {
        long startTime = System.nanoTime();

        int M = (int) (1e9 + 7);

        int n = nextInt();
        int m = nextInt();
        char[] c = nextToken().toCharArray();
        int[] s = new int[n + 1];
        for (int i = 0; i < n; i++) {
            s[i + 1] = s[i] + (c[i] - '0');
        }

        for (int j = 0; j < m; j++) {
            int a = nextInt() - 1;
            int b = nextInt();
            int x = s[b] - s[a];
            int y = b - a - x;
            println((get(2, x, M) + M - 1) % M * get(2, y, M) % M);
        }

        if (fileIOMode) {
            System.out.println((System.nanoTime() - startTime) / 1e9);
        }
        out.close();
    }

    private long get(long x, int pow, int m) {
        if (pow == 0) {
            return 1;
        }
        if ((pow & 1) == 0) {
            return get(x * x % m, pow >> 1, m);
        } else {
            return x * get(x, pow - 1, m) % m;
        }
    }

    //-----------------------------------------------------------------------------------

    private static boolean fileIOMode;
    private static BufferedReader in;
    private static PrintWriter out;
    private static StringTokenizer tokenizer;

    public static void main(String[] args) throws Exception {
        fileIOMode = args.length > 0 && args[0].equals("!");
        if (fileIOMode) {
            in = new BufferedReader(new FileReader("a.in"));
            out = new PrintWriter("a.out");
        } else {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        }
        tokenizer = new StringTokenizer("");

        new Thread(new A()).start();
    }

    private static String nextLine() {
        try {
            return in.readLine();
        } catch (IOException e) {
            return null;
        }
    }

    private static String nextToken() {
        while (!tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(nextLine());
        }
        return tokenizer.nextToken();
    }

    private static int nextInt() {
        return Integer.parseInt(nextToken());
    }

    private static long nextLong() {
        return Long.parseLong(nextToken());
    }

    private static double nextDouble() {
        return Double.parseDouble(nextToken());
    }

    private static BigInteger nextBigInteger() {
        return new BigInteger(nextToken());
    }

    private static void print(Object o) {
        if (fileIOMode) {
            System.out.print(o);
        }
        out.print(o);
    }

    private static void println(Object o) {
        if (fileIOMode) {
            System.out.println(o);
        }
        out.println(o);
    }

    private static void printf(String s, Object... o) {
        if (fileIOMode) {
            System.out.printf(s, o);
        }
        out.printf(s, o);
    }
}