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 javafx.util.Pair;

import java.io.*;
import java.math.*;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException {
        PrintWriter out = new PrintWriter(System.out);
        Main mm = new Main();
        mm.problemC(new Input(new BufferedReader(new InputStreamReader(System.in))), out);
        out.close();
    }

    static void problemC(Input in, PrintWriter out) throws IOException {
        int n=in.nextInt(),q=in.nextInt();
        int md=1000000007;
        int[] ar=new int[n+1],pw=new int[n+1];
        for (int i = 0; i < n; i++) {
            ar[i+1]=ar[i]+in.nextChar()-'0';
        }
        pw[0]=1;
        for (int i = 1; i <= n; i++) {
            pw[i]=(pw[i-1]<<1)%md;
        }
        for (int i = 0; i < q; i++) {
            int l=in.nextInt(),r=in.nextInt();
            int len=r-l+1;
            int zeros=len-(ar[r]-ar[l-1]);
            int res=(pw[len]-pw[zeros]+md)%md;
            out.println(res);
        }
    }


    static void problemB(Input in, PrintWriter out) throws IOException {
        int n=in.nextInt(),ans=1,mx=0;
        boolean bl=false;
        int pow=0;
        if(n==1)
            out.println("1 0");
        else{
            eratosfen e=new Main().new eratosfen(n);
            ArrayList<Integer> a=e.fillSieve();
            for (int i = 0; n!=1; i++) {
                int b=a.get(i);
                int k=0;
                while (n%b==0){
                    n/=b;
                    k++;
                }
                if(k!=0) {
                    if(mx!=0&&k!=mx)bl=true;
                    mx=Math.max(mx,k);
                    ans*=b;
                }
            }
            int r=1;
            while(r<mx) {
                r*=2;
                pow++;
            }
            if(r!=mx||bl)pow++;
            out.print(ans);out.print(' ');out.println(pow);
        }

    }

    static void problemA(Input in, PrintWriter out) throws IOException {
        int n=in.nextInt();
        int a=in.nextInt(),b=a,k=1,d=1,el=a,mx=0;
        for (int i = 1; i < n; i++) {
            a=in.nextInt();
            if (a==b+1) {
                d++;
                if (d >= k) {
                    k = d;
                    el = a;
                }
            }else{
                d = 1;
                if (k == b) {
                    mx = b - 1;
                }
            }
            b=a;
        }
        if (b == k) {
            mx = k - 1;
        }
        mx=Math.max(mx,(el==1000)?d-1:0);
        mx=Math.max(mx,k-2);
        out.println(mx);
    }

    public class eratosfen {
        boolean[] primes;
        public eratosfen(int n) {
            primes=new boolean[n+1];
        }
        public ArrayList<Integer> fillSieve() {
            ArrayList<Integer> a=new ArrayList<>();
            Arrays.fill(primes, true);
            primes[0] = false;
            primes[1] = false;
            for (int i = 2; i < primes.length; ++i) {
                if (primes[i]) {
                    for (int j = 2; i * j < primes.length; ++j) {
                        primes[i * j] = false;
                    }
                    a.add(i);
                }
            }
            return a;
        }
    }

    static class Input {
        BufferedReader in;
        StringBuilder sb = new StringBuilder();

        public Input(BufferedReader in) {
            this.in = in;
        }

        public Input(String s) {
            this.in = new BufferedReader(new StringReader(s));
        }

        public String next() throws IOException {
            sb.setLength(0);
            while (true) {
                int c = in.read();
                if (c == -1) {
                    return null;
                }
                if (" \n\r\t".indexOf(c) == -1) {
                    sb.append((char) c);
                    break;
                }
            }
            while (true) {
                int c = in.read();
                if (c == -1 || " \n\r\t".indexOf(c) != -1) {
                    break;
                }
                sb.append((char) c);
            }
            return sb.toString();
        }

        public char nextChar() throws IOException {
            while (true) {
                int c = in.read();
                if (c == -1) {
                    return (char)c;
                }
                if (" \n\r\t".indexOf(c) == -1) {
                    return (char)c;
                }
            }
        }

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

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

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }
    }
}