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.*;

public class Banhmi {
    
    
    
    /************************ SOLUTION STARTS HERE ************************/
    
    static class MM {       // MM (Modular Math) class 
        static final long mod = (long) (1e9) + 7; // Default
        static long sub(long a, long b) {return (a - b  + mod) % mod;}
        static long mul(long a, long b) {return ((a % mod) * (b % mod)) % mod;}
        static long add(long a, long b) {return (a + b) % mod;}
        static long div(long a, long b) {return mul(a, modInverse(b));}
        static long modInverse(long n)  {return modPow(n, mod - 2);} // Fermat's little theorem
        static long modPow(long a , long b) {
            long pow = 1;
            while(b > 0) {
                if((b & 1L) == 1)
                    pow = ((pow * a) % mod);

                a = ((a * a) % (mod));
                b >>= 1;
            }
            return pow;
        }
    }
    
    private static void solve() {
        
        int n = nextInt();
        int q = nextInt();
        
        int val[] = new int[n + 1];
        char str[] = nextLine().toCharArray();
        for(int i = 0; i < n; i++) 
            val[i + 1] = val[i] + (str[i] - '0');
        
        while(q-->0) {
            int l = nextInt();
            int r = nextInt();
            int len = r - l + 1;
            int ones = val[r] - val[l - 1];
            int zeroes = len - ones;
            long conjMax = MM.mul(MM.sub(MM.modPow(2, ones), 1), MM.modPow(2, zeroes));
            
            println(conjMax);
        }
    }
    
    
    
    /************************ SOLUTION ENDS HERE ************************/
    
    
    
    
    
    /************************ TEMPLATE STARTS HERE **********************/
    
    public static void main(String[] args) throws IOException {
        reader = new BufferedReader(new InputStreamReader(System.in));
        writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false);
        st     = null;
        solve();
        reader.close();
        writer.close();
    }
    
    static BufferedReader reader;
    static PrintWriter    writer;
    static StringTokenizer st;
    
    static String next()
    {while(st == null || !st.hasMoreTokens()){try{String line = reader.readLine();if(line == null){return null;}            
    st = new StringTokenizer(line);}catch (Exception e){throw new RuntimeException();}}return st.nextToken();}
    static String nextLine()  {String s=null;try{s=reader.readLine();}catch(IOException e){e.printStackTrace();}return s;}             
    static int    nextInt()   {return Integer.parseInt(next());}
    static long   nextLong()  {return Long.parseLong(next());}     
    static double nextDouble(){return Double.parseDouble(next());}
    static char   nextChar()  {return next().charAt(0);}
    static int[]  nextIntArray(int n)         {int[] a= new int[n];   int i=0;while(i<n){a[i++]=nextInt();}  return a;}
    static long[] nextLongArray(int n)        {long[]a= new long[n];  int i=0;while(i<n){a[i++]=nextLong();} return a;}    
    static int[]  nextIntArrayOneBased(int n) {int[] a= new int[n+1]; int i=1;while(i<=n){a[i++]=nextInt();} return a;}            
    static long[] nextLongArrayOneBased(int n){long[]a= new long[n+1];int i=1;while(i<=n){a[i++]=nextLong();}return a;}            
    static void   print(Object o)  { writer.print(o);  }
    static void   println(Object o){ writer.println(o);}
    
    /************************ TEMPLATE ENDS HERE ************************/
    
}