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:
//package ;
import java.io.*;
import java.util.*;

public class C
{
	static int freq[][],n;
	public static void main(String[] args) throws IOException 
	{
		Scanner sc = new Scanner();
		PrintWriter pw = new PrintWriter(System.out);
		n=sc.nextInt();int q=sc.nextInt();
		freq=new int[2][n];
		char c[]=sc.nextLine().toCharArray();
		for(int i=0;i<n;i++)
		{
			if(i==0)
			{
				if(c[i]=='0')
					freq[0][i]=1;
				else
					freq[1][i]=1;	
			}
			else
			{
				if( c[i]=='0')
				{
					freq[0][i]=freq[0][i-1]+1;
					freq[1][i]=freq[1][i-1];	
				}
				else
				{	
					freq[0][i]=freq[0][i-1];				
					freq[1][i]=freq[1][i-1]+1;	
				}
			}
		}
		while(q-->0)
		{
			int l=sc.nextInt()-1,r=sc.nextInt()-1;
			long ans;
			if(l==0)
			{
				int ones=freq[1][r];
				int zero=freq[0][r];
				
				long sum_1=pow_mod(2, ones,mod)-1;
				long sum_z=(sum_1*(pow_mod(2, zero,mod)-1))%mod;
				ans=sum_1+sum_z;
			}
			else
			{
				int ones=freq[1][r]-freq[1][l-1];
				int zero=freq[0][r]-freq[0][l-1];
				
				long sum_1=pow_mod(2, ones,mod)-1;
				long sum_z=(sum_1*(pow_mod(2, zero,mod)-1))%mod;
				ans=sum_1+sum_z;
			}
			pw.println(ans%mod);
		}
		pw.close();    
	}
	public static long pow_mod(long b,long p,long m)//O(log(p))
	{
		long res=1;
		while(p>0)
		{
			if(p%2==1)
				res=res*b%m;
			b=b*b%m;
			p/=2;
		}
		return res;
	}
	static long mod=(long)1e9+7;
	static class Scanner {
		BufferedReader br;
		StringTokenizer st;
		
		Scanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		String next() throws IOException {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(br.readLine());
			}
			return st.nextToken();
		}
		
		int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
		
		long nextLong() throws IOException {
			return Long.parseLong(next());
		}
		
		double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
		
		String nextLine() throws IOException {
			return br.readLine();
		}
		boolean hasnext() throws IOException{
			return br.ready();
		}
		
	}
}