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.math.BigInteger;
import java.util.StringTokenizer;

public class Banhi {
    static long modPow(int a, int e, int mod)
    {
        a %= mod;
        long res = 1;
        while(e > 0)
        {
            if((e & 1) == 1)
                res =(1l*res * a) % mod;
            a = (int)((1l*a * a) % mod);
            e >>= 1;
        }
        return res;
    }


    public  static  void  main(String[]args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int n = sc.nextInt();
        int q = sc.nextInt();
        zz []l= new zz[n];
        String s= sc.nextLine();
        int o=0;
        int z =0;
        for (int i =0;i<n;i++){
            if (s.charAt(i)=='1')
                o++;
            else z++;
            l[i]=new zz(o,z);
        }
        while (q-->0){
            int a = sc.nextInt()-2;
            int b = sc.nextInt()-1;
            long ans =0;
         if (a<0)
            ans=(modPow(2,l[b].o,1000000007)-1)*modPow(2,l[b].z,1000000007);
         else
             ans=(modPow(2,l[b].o-l[a].o,1000000007)-1)*modPow(2,l[b].z-l[a].z,1000000007);
            out.println(ans<0?0:ans%1000000007);
        }
        out.flush();
    }
    static class zz{
        int o ;
        int z;
        zz(int o ,int z){
            this.o=o;
            this.z=z;
        }
    }
    static class Scanner
    {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

        public String next() throws IOException
        {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

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

        public long nextLong() throws IOException {return Long.parseLong(next());}

        public String nextLine() throws IOException {return br.readLine();}

        public double nextDouble() throws IOException
        {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if(x.charAt(0) == '-')
            {
                neg = true;
                start++;
            }
            for(int i = start; i < x.length(); i++)
                if(x.charAt(i) == '.')
                {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                }
                else
                {
                    sb.append(x.charAt(i));
                    if(dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg?-1:1);
        }

        public boolean ready() throws IOException {return br.ready();}


    }
}