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.*;

public class ProblemC {
    public static final long M = 1_000_000_000 + 7;


    /* Iterative Function to calculate (x^y)%p in O(log y) */
    static long power(long x, long y, long p) {
        long res = 1;
        x = x % p;

        while (y > 0) {
            if ((y & 1l) > 0) res = (res * x) % p;
            y = y >> 1; // y = y/2
            x = (x * x) % p;
        }
        return res;
    }

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        /* INPUT */
        int n = scan.nextInt(), q = scan.nextInt();
        String s = scan.next();

        int[] accu = new int[n];
        int curr = 0;
        for (int i = 0; i < n; i++) {
            int ai = s.charAt(i) - '0';
            curr += ai;
            accu[i] = curr;
        }
        // System.out.println(Arrays.toString(accu));

        long[] pow2 = new long[n + 1];
        pow2[0] = 1;
        long p2 = 1;
        for (int i = 1; i <= n; i++) {
            p2 *= 2;
            p2 %= M;
            pow2[i] = p2;
        }
        // System.out.println(Arrays.toString(pow2));


        StringBuilder res = new StringBuilder();
        while (q-- > 0) {
            int l = scan.nextInt() - 1, r = scan.nextInt() - 1;
            int len = r - l + 1;
            int ones = accu[r];
            if (l > 0) ones -= accu[l - 1];
            int zero = len - ones;

            // System.out.println(ones + " 1:0 " + zero);

            long onesMax = power(2l, ones - 1, M);
            // System.out.println(onesMax);

            // a * (1 - r^n)/ ( 1 - r)
            // r = 2
            long onesSum = pow2[ones]; //power(2l, ones, M);
            onesSum--;
            if (onesSum <= 0) onesSum += M;

            // onesSum * (1 - r ^2) / -1
            long zeroSum = pow2[zero]; //power(2l, zero, M);
            zeroSum--;
            if (zeroSum <= 0) onesSum += M;
            zeroSum *= onesSum;
            zeroSum %= M;

            res.append(((onesSum + zeroSum) % M)).append("\n");
        }
        System.out.println(res);
    }
}