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.*;
import java.util.*;
public class Main {
    public static void main(String args[])
    {
        FastReader input=new FastReader();
        PrintWriter out=new PrintWriter(System.out);
        int T=1;
        while(T-->0)
        {
            int n=input.nextInt();
            int q=input.nextInt();
            String s=input.next();
            Pair p[]=new Pair[n];
            int z=0,o=0;
            for(int i=0;i<n;i++)
            {
                if(s.charAt(i)=='1')
                {
                    o++;
                }
                else
                {
                    z++;
                }
                p[i]=new Pair(z,o);
            }
            for(int i=0;i<q;i++)
            {
                int l=input.nextInt();
                int r=input.nextInt();
                l--;r--;
                if(l==0)
                {
                    z=p[r].zero;
                    o=p[r].one;
                }
                else
                {
                    z=p[r].zero-p[l-1].zero;
                    o=p[r].one-p[l-1].one;
                }
                if(o==0)
                {
                    out.println(0);
                }
                else
                {
                    long m=1000000007;
                    long s1=power(2,o,m);
                    s1=(s1-1)%m;
                    long f=power(2,o,m);
                    f=(f-1)%m;
                    long s2=power(2,z,m);
                    s2=(s2-1)%m;
                    s2=(s2*f)%m;
                    s1=(s1+s2)%m;
                    out.println(s1);
                }
            }
        }
        out.close();
    }
    public static long power(long a,long b, long m)
    {
        long res=1;
        while(b>0)
        {
            if(b%2!=0)
            {
                res=(res%m*a%m)%m;
            }
            b=b/2;
            a=(a%m*a%m)%m;
        }
        return res;
    }
    static class Pair
    {
        int zero,one;
        public Pair(int zero,int one)
        {
            this.one=one;
            this.zero=zero;
        }
    }
    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();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {

            return Long.parseLong(next());
        }

        double nextDouble()
        {

            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
}