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.*;
import java.io.*;
import java.math.*;
 
public class Main
{
    static void precompute_pow(long arr[],int n){
        arr[0]=1l;

        for(int i=1;i<=n;i++) arr[i]=(arr[i-1]*2l)%mod;
    }

    public static void process()throws IOException
    {
        int n=ni(),q=ni();
        long pow[]=new long[n+1];
        char arr[]=(" "+nln()).toCharArray();

        precompute_pow(pow,n);

        int pref_1[]=new int[n+1],pref_0[]=new int[n+1];

        for(int i=1;i<=n;i++){
            pref_1[i]=pref_1[i-1]+(arr[i]-'0');
            pref_0[i]=pref_0[i-1]+(1-(arr[i]-'0'));
        }

        while(q-- > 0){
            int l=ni(),r=ni();
            long res=pow[pref_1[r]-pref_1[l-1]]-1l;
            res%=mod;

            res=res+res*(pow[pref_0[r]-pref_0[l-1]]-1l)%mod;
            res%=mod;

            pn(res);
        }
    }
 
 
    static FastReader sc;
    static PrintWriter out;
    public static void main(String[]args)throws IOException
    {
        out = new PrintWriter(System.out);
        sc=new FastReader();
 
        long s = System.currentTimeMillis();
        int t=1;
        //t=ni();
        while(t-->0)
            process();
 
        out.flush();
        System.err.println(System.currentTimeMillis()-s+"ms");
    }
    
    
    static void pn(Object o){out.println(o);}
    static void p(Object o){out.print(o);}
    static int ni()throws IOException{return Integer.parseInt(sc.next());}
    static long nl()throws IOException{return Long.parseLong(sc.next());}
    static double nd()throws IOException{return Double.parseDouble(sc.next());}
    static String nln()throws IOException{return sc.nextLine();}
    static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int bit(long n)throws IOException{return (n==0)?0:(1+bit(n&(n-1)));}
    
    static long mod=(long)1e9+7l;
 
    static<T> void r_sort(T arr[],int n){
        Random r = new Random(); 
        for (int i = n-1; i > 0; i--){ 
            int j = r.nextInt(i+1); 
                  
            T temp = arr[i]; 
            arr[i] = arr[j]; 
            arr[j] = temp; 
        } 
        Arrays.sort(arr); 
    }
    
/////////////////////////////////////////////////////////////////////////////////////////////////////////
    static class FastReader{ 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader(){ 
            br = new BufferedReader(new InputStreamReader(System.in)); 
        } 
  
        String next(){ 
            while (st == null || !st.hasMoreElements()){ 
                try{ st = new StringTokenizer(br.readLine()); } catch (IOException  e){ e.printStackTrace(); } 
            } 
            return st.nextToken(); 
        } 
  
        String nextLine(){ 
            String str = ""; 
            try{ str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } 
            return str; 
        } 
    } 
/////////////////////////////////////////////////////////////////////////////////////////////////////////////
}