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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.BitSet;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Set;
import java.util.StringTokenizer;

public class q5 {
	static int[] visited;
	 public static int gcd(int a,int b) {
		 if(b==0) return a;
		 else return gcd(b,a%b);
	 }
	 // Credits TO GFG
	 static long power(long x, long y, long p) 
	    { 
	        // Initialize result 
	        long res = 1;      
	         
	        // Update x if it is more   
	        // than or equal to p 
	        x = x % p;  
	      
	        while (y > 0) 
	        { 
	            // If y is odd, multiply x 
	            // with result 
	            if((y & 1)==1) 
	                res = (res * x) % p; 
	      
	            // y must be even now 
	            // y = y / 2 
	            y = y >> 1;  
	            x = (x * x) % p;  
	        } 
	        return res; 
	    } 
	 
	
	 
	 
	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		Reader.init(System.in);
		int n=Reader.nextInt();
		int q=Reader.nextInt();
		String s=Reader.next();
		int[] arr=new int[n+1];
		long mod=1000000007;
		for(int i=0;i<n;i++) {
			if(s.charAt(i)=='1') {
				arr[i+1]=arr[i]+1;
			}
			else {
				arr[i+1]=arr[i];
			}
		}
		for(int i=0;i<q;i++) {
			int l=Reader.nextInt();
			int r=Reader.nextInt();
			int length=r-l+1;
			int no=arr[r]-arr[l-1];
			int nz=length-no;
			long val=q5.power(2, no, mod)-1;
			long tv=q5.power(2, nz, mod)-1;
			long ans=(val*tv)%mod+val;
			ans%=mod;
			System.out.println(ans);
			
			
		}
		
				
			
	}
}
class Node2{
	int edge1;
	int edge2;
	long weight;
	int number;
	Node2(int a,int b,long c,int d){
		edge1=a;edge2=b;
		weight=c;number=d;
	}
}
 
 
 class DisJoint {
		int[] parent;
		int[] rank;
		int[] size;
		DisJoint(int n){
			
			parent=new int[n+1];
			rank=new int[n+1];
			size=new int[n+1];
			for(int i=0;i<=n;i++) {
				parent[i]=i;
				size[i]=1;
			}
		}

		int find(int value) {
			int par=parent[value];
			if(par==value)
				return par;
			parent[value]=find(par);
			return parent[value];
		}
		
		void union(int data1,int data2) {
			int parent1=find(data1);
			int parent2=find(data2);
			if(parent1!=parent2) {
				if(rank[parent1]>=rank[parent2]) {
					parent[parent2]=parent1;
					if(rank[parent1]==rank[parent2])
						rank[parent1]++;
					size[parent1]+=size[parent2];
				}
				else {
					parent[parent1]=parent2;	
					size[parent2]+=size[parent1];
		}

		}
		}
}

class Reader {
    static BufferedReader reader;
    static StringTokenizer tokenizer;
 
    /** call this method to initialize reader for InputStream */
    static void init(InputStream input) {
        reader = new BufferedReader(
                     new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }
 
    /** get next word */
    
    static String next() throws IOException {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                   reader.readLine() );
        }
        return tokenizer.nextToken();
    }
 
    static int nextInt() throws IOException {
        return Integer.parseInt( next() );
    }
    static long nextLong() throws IOException {
        return Long.parseLong( next() );
    }
    static double nextDouble() throws IOException {
        return Double.parseDouble( next() );
    }
}