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.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigDecimal;
import java.util.*;

public class C {

    private String readLine() {
        StringBuilder sb = new StringBuilder();
        int b = 0;
        while (b != '\n' && b >= 0) {
            try {
                b = System.in.read();
            } catch (IOException e) {
                return null;
            }
            if (b != '\r' && b != '\n' && b >= 0) {
                sb.append((char) b);
            }
        }
        if (sb.length() == 0 && b < 0) {
            return null;
        }
        return sb.toString().trim();
    }

    private int parseInt(String s) {
        int result = 0;
        int sign = (s.charAt(0) == '-') ? -1 : 1;
        if (sign == -1) {
            s = s.substring(1);
        }
        if (s.charAt(0) == '+') {
            s = s.substring(1);
        }
        int i = 0, max = s.length();
        if (max > 0) {
            while (i < max) {
                result *= 10;
                result += s.charAt(i++) - 48;
            }
        }
        return sign * result;
    }

    private void work() {
        p2 = new long[100010];
        p2[0] = 1;
        for (int i = 1; i < p2.length; i++) {
            p2[i] = (2 * p2[i - 1]) % MOD;
        }

        String[] spl = readLine().split("\\s+");
        int n = parseInt(spl[0]);
        int q = parseInt(spl[1]);
        char[] s = readLine().trim().toCharArray();
        int[] a = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            a[i] = a[i - 1];
            if (s[i - 1] == '1') a[i]++;
        }

        while (q-- > 0) {
            spl = readLine().split("\\s+");
            int l = parseInt(spl[0]);
            int r = parseInt(spl[1]);
            int ones = a[r] - a[l - 1];
            int zeros = r - l + 1 - ones;
            System.out.println(f(ones, zeros));
        }
    }

    private static final long MOD = 1000000007;
    private long[] p2;

    private long f(int ones, int zeros) {
        if (ones == 0) return 0;
        return ((p2[ones] - 1) * p2[zeros]) % MOD;
    }

    public static void main(String[] args) {
        new C().work();
    }
}
