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.util.Arrays;
import java.util.Scanner;

/**
 * Created by Jamie on 2/14/2019.
 */
public class Bahnmi {

    public static int MODULUS = 1_000_000_007;

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        FTree ftree = new FTree(100_000);

        int n = scan.nextInt();
        int q = scan.nextInt();

        scan.nextLine();
        String temp = scan.nextLine();
        int[] bahn = new int[n];
        for (int i = 0; i < n; i++) {
            bahn[i] = temp.charAt(i) - '0';
            if (bahn[i] == 1) {
                ftree.update(i+1, 1);
            }
        }

        int total1s = ftree.getSum(n);
        long[] partials = new long[total1s + 1];
        partials[0] = 0;
        for (int i = 1; i < total1s + 1; i++) {
            long val = (partials[i-1] + 1) * 2 - 1;
            partials[i] = val % MODULUS;
        }
        //System.out.println(Arrays.toString(partials));

        long[] pow2s = new long[n - total1s + 1];
        pow2s[0] = 1;
        for (int i = 1; i < n-total1s + 1; i++) {
            pow2s[i] = (pow2s[i-1] * 2) % MODULUS;
        }

        for (int j = 0; j < q; j++) {
            int l = scan.nextInt();
            int r = scan.nextInt();
            int num1s = ftree.getRange(l, r);
            if (num1s < 0) {
                System.out.println("Overflow " + temp);
                System.exit(0);
            }
            int size = r-l+1;

//            long total = 0;
            long total = partials[num1s];

            //Double for every 0
            total *= pow2s[size-num1s];
            total %= MODULUS;
            System.out.println(total);
        }

    }

    static class FTree {

        //1 index fenwick tree
        private int[] tree;
        private int size;

        public FTree(int n) {
            size = n+1;
            tree = new int[size];
        }

        //Get range with inclusive boundaries
        public int getRange(int a, int b) {
            return getSum(b) - getSum(a-1);
        }


        public void update(int pos, int val) {
            while(pos < size) {
                tree[pos] += val;
                pos += Integer.lowestOneBit(pos);
            }
        }

        public int getSum(int pos) {
            int retVal = 0;
            while (pos >= 1) {
                retVal += tree[pos];
                pos -= Integer.lowestOneBit(pos);
            }
            return retVal;
        }
    }
}