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.Scanner;

public class c {
	static long oo = 1_000_000_007;
	
	
	static long add(long a, long b) {
		return (a+b)%oo;
	}
	static long negate(long a) {
		return (oo-a)%oo;
	}
	static long mult(long a, long b) {
		return (a*b)%oo;
	}
	static long pow(long a, long p) {
		if(p == 0) return 1;
		if(p == 1) return a;
		long rec = pow(a, p/2);
		rec = mult(rec, rec);
		if(p % 2 == 0) return rec;
		rec = mult(rec, a);
		return rec;
	}
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int N = in.nextInt();
		int Q = in.nextInt();
		char[] str = in.next().toCharArray();
		
		long[] prefsum0 = new long[N+1];
		long[] prefsum1 = new long[N+1];
		for(int n=0;n<N;n++) {
			prefsum0[n+1] = prefsum0[n] + (str[n] == '0'? 1: 0);
			prefsum1[n+1] = prefsum1[n] + (str[n] == '1'? 1: 0);
		}
		
		StringBuilder sb = new StringBuilder();
		for(int q=0;q<Q;q++) {
			int l = in.nextInt()-1;
			int r = in.nextInt();
			
			long num0 = prefsum0[r] - prefsum0[l];
			long num1 = prefsum1[r] - prefsum1[l];
			
			long partOne = add(pow(2L, num1), negate(1));
			long partTwo = mult(partOne, add(pow(2L, num0), negate(1)));
			sb.append(add(partOne, partTwo)).append('\n');
		}
		
		System.out.println(sb);
	}
}