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.util.*;
import java.io.*;

public class Banhmi {

	// https://codeforces.com/contest/1062/problem/C
	
	public static void main(String[] args) throws IOException, FileNotFoundException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		//BufferedReader in = new BufferedReader(new FileReader("Banhmi"));

		StringTokenizer st = new StringTokenizer(in.readLine());
		int n = Integer.parseInt(st.nextToken());
		int q = Integer.parseInt(st.nextToken());
		char[] arr = new char[n];
		arr = in.readLine().toCharArray();
		int[] one_count = new int[n+1];
		for (int i=0; i<n; i++) {
			one_count[i+1] = one_count[i];
			if (arr[i] == '1') one_count[i+1]++;
		}
		
		long mod = (long)1e9 + 7;
		long[] powers_of_two = new long[n+10];
		powers_of_two[0] = 1;
		for (int i=1; i<powers_of_two.length; i++) {
			powers_of_two[i] = powers_of_two[i-1] * 2;
			powers_of_two[i] %= mod;
		}
		
		for (int i=0; i<q; i++) {
			st = new StringTokenizer(in.readLine());
			int l = Integer.parseInt(st.nextToken());
			int r = Integer.parseInt(st.nextToken());
			int num_ones = one_count[r] - one_count[l-1];
			int num_zeroes = r-l+1 - num_ones;
			long curans=0;
			curans += powers_of_two[num_ones]-1;
			curans %= mod;
			long others = curans;
			others *= powers_of_two[num_zeroes]-1;
			others %= mod;
			curans += others;
			curans = (curans%mod+mod)%mod;
			
			System.out.println(curans);
		}
	}
}
