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.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;

public class Solution {

    // Complete the sockMerchant function below.
  

    private static final Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) throws IOException {
        int MAX = 100005, MOD = 1000000007;
        long[] power2 = new long[MAX];
        power2[0] = 1;
        for(int i = 1; i < MAX; i++){
            power2[i] = power2[i - 1] * 2;
            power2[i] %= MOD;
        }
        int l, r, a, b;
        int[] count = new int[MAX];
        int n = scanner.nextInt(), q = scanner.nextInt();
        String s = scanner.next();
        for(int i = 0; i < s.length(); i++){
            if(i > 0)count[i] = count[i - 1];
            if(s.charAt(i) == '1'){
                count[i]++;
            }
        }
        for(int i = 0; i < q; i++){
            l = scanner.nextInt();
            r = scanner.nextInt();
            l--; r--;
            b = count[r] - (l > 0 ? count[l - 1] : 0);
            a = r - l + 1 - b;
            // System.out.println(a + " " + b);
            // System.out.println(power(2, 10, MOD));
            // System.out.println(power(2, 10000, MOD));
            // System.out.println(power(2, a + b, MOD));
            long ans;
            if(b == 0)ans = 0;
            else ans = (power2[a + b] - power2[a]) % MOD;
            if(ans < 0)ans += MOD;
            System.out.println(ans);
        }
    }
}