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.Scanner;

public class C {
	
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int q = scan.nextInt();
		String S = scan.next();
		int[] ones = new int[n+1];
		int[] zeros = new int[n+1];
		int numOfOnes = 0;
		int numOFZeros = 0;
		for (int i = 0 ; i < n ; i++) {
			if (S.charAt(i) == '1') {
				numOfOnes++;
				ones[i+1] = numOfOnes;
				zeros[i+1] = numOFZeros;
			} else {
				numOFZeros++;
				zeros[i+1] = numOFZeros;
				ones[i+1] = numOfOnes;
			}
		}
		
		long[] pow = new long[100001];
		pow[0] = 1;
		
		for (int i = 1 ; i < 100001 ; i++) {
			pow[i] = (pow[i-1] * 2) % 1000000007;
		}
		

		for (int i = 0 ; i < q ; i++) {
			int l = scan.nextInt();
			int r = scan.nextInt();
			
			int numOfOne = ones[r] - ones[l-1];
			int numOfZero = zeros[r] - zeros[l-1];
			
			long first = pow[numOfOne]-1;
			
//			System.out.println(first);
			long second = pow[numOfOne + numOfZero]-1;
//			System.out.println(second);
			long third = pow[numOfOne]-1;
			long fourth = pow[numOfZero]-1;
//			System.out.println(fourth);
//			System.out.println(third);
			if (numOfZero > 0) {
				third = (third + fourth + 1000000007) % 1000000007;
				second -= third;
				first = (first + second + 1000000007) % 1000000007;	
				System.out.println(first % 1000000007);
			} else {
			
				System.out.println(first % 1000000007);
			
			}
			
			
		}
	}
	

	static long power(long x,  
            long y, long p) 
{ 
long res = 1; // Initialize result 

// Update x if it is more  
// than or equal to p 
x = x % p;  

while (y > 0)  
{ 
  // If y is odd, multiply  
  // x with the result 
  if ((y & 1) > 0) 
      res = (res * x) % p; 

  // y must be even now 
  y = y >> 1; // y = y/2 
  x = (x * x) % p; 
} 
return res; 
} 


}
