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 cfc {
    static final int MOD=(int)(1e9+7);
    public static void main(String[] args) {
        FastScanner scan=new FastScanner();
        PrintWriter out=new PrintWriter(System.out);
        int n=scan.nextInt(), q=scan.nextInt();
        exp=new long[n+1];
        exp[0]=1;
        for(int i=1;i<exp.length;i++) exp[i]=(exp[i-1]<<1)%MOD;
        String s=scan.next();
        int[] ones=new int[n];
        for(int i=0;i<n;i++) {
            int add=i==0?0:ones[i-1];
            ones[i]=s.charAt(i)=='1'?1+add:0+add;
        }
//        System.out.println(Arrays.toString(exp));
        for(int i=0;i<q;i++) {
            int l=scan.nextInt()-1, r=scan.nextInt()-1;
            int one=ones[r];
            if(l>0) one-=ones[l-1];
            int zero=r-l+1-one;
//            System.out.println(zero+" "+one);
            long res=((long)exp[one]-1+MOD)%MOD;
            res=(res*(long)exp[zero])%MOD;
            // System.out.println(one+" "+zero);
            out.println(res);
        }
        out.close();
    }
    static long[] exp;
    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public FastScanner() {
            try {
                br = new BufferedReader(new InputStreamReader(System.in));
                st = new StringTokenizer(br.readLine());
            } catch (Exception e){e.printStackTrace();}
        }
 
        public String next() {
            if (st.hasMoreTokens()) return st.nextToken();
            try {st = new StringTokenizer(br.readLine());}
            catch (Exception e) {e.printStackTrace();}
            return st.nextToken();
        }
 
        public int nextInt() {return Integer.parseInt(next());}
 
        public long nextLong() {return Long.parseLong(next());}
 
        public double nextDouble() {return Double.parseDouble(next());}
 
        public String nextLine() {
            String line = "";
            if(st.hasMoreTokens()) line = st.nextToken();
            else try {return br.readLine();}catch(IOException e){e.printStackTrace();}
            while(st.hasMoreTokens()) line += " "+st.nextToken();
            return line;
        }
    }
}