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

public class C {

	public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

	public static void main(String[] args) throws IOException {
		readInput();
		out.close();
	}

	public static void readInput() throws IOException {
		// br = new BufferedReader(new FileReader(".in"));
		// out = new PrintWriter(new FileWriter(".out"));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		int q = Integer.parseInt(st.nextToken());
		int[][] p = new int[n+1][2];
		String s = br.readLine();
		for (int i = 0; i < n; i++) {
			if (s.charAt(i) == '1') {
				p[i+1][1] = 1;
			}
			else p[i+1][0] = 01;
			
			p[i+1][0] += p[i][0];
			p[i+1][1] += p[i][1];
		}
		while(q-->0) {
			st = new StringTokenizer(br.readLine());
			int l = Integer.parseInt(st.nextToken());
			int r = Integer.parseInt(st.nextToken());
			long res = pow(p[r][1]-p[l-1][1])-1 + mod;
			res %= mod;
			res *= pow(p[r][0] - p[l-1][0]);
			out.println(res % mod);
		}
	}
	static long mod = (int)1e9+7;
	static long pow(int e) {
		if (e == 0) return 1;
		long r = pow(e/2);
		r = r*r%mod;
		if (e % 2 ==1) return 2 * r % mod;
		return r;
	}
}