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.*;

public class segment_tree{
	
	static int max = (int)1e5+5;
	static int[] a = new int[2*max];
	static long mod = (long)1e9+7;

	static void build(int n){
		for(int i=n-1;i>0;i--)a[i] = a[i<<1]+a[i<<1|1];
	}

	static int query(int l,int r,int n){
		int res = 0;
		for(l = l+n,r = r+n;l<r;l>>=1,r>>=1){
			if((l&1)==1)res+=a[l++];
			if((r&1)==1)res+=a[--r];
		}
		return res;
	}

	public static void main(String[] args){
		Scanner in = new Scanner(System.in);
		int n = in.nextInt(), q = in.nextInt();
		String s = in.next();
		for(int i=0;i<n;i++)a[i+1] = (s.charAt(i)-'0')+a[i];
		// build(n);

		long[] p =new long[n+1];
		p[0] = 1L;
		for(int i=1;i<=n;i++)p[i] = (2L*p[i-1])%mod;
		//System.out.println(p[n]);
		while(q-->0){
			int l = in.nextInt(),r = in.nextInt();
			int n1 = a[r]-a[l-1];//query(l-1,r,n);
			//System.out.println((p[r-l+1]*(p[n1]-1L))/p[n1]);
			System.out.println(((p[n1]-1)*p[r-l+1-n1])%mod);
		}
	}
}