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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class C {

    static int mod = (int)1e9 + 7;
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        PrintWriter pw = new PrintWriter(System.out);

        int n = sc.nextInt();
        int q = sc.nextInt();
        int[] pow2 = new int[1000000];
        pow2[0] = 1;
        for (int i = 1; i < pow2.length; i++) {
            pow2[i] = (int)((1l*pow2[i-1]*2)%mod);
        }

        char[] arr = sc.next().toCharArray();
        int[] cumOnes = new int[n];
        int[] cumZeros = new int[n];
        for (int i = 0; i < n; i++) {
            if(arr[i] == '1')
                cumOnes[i] = 1;
            else
                cumZeros[i] = 1;
            if(i > 0) {
                cumOnes[i] += cumOnes[i - 1];
                cumZeros[i] += cumZeros[i-1];
            }
        }

        while(q-->0)
        {
            int l = sc.nextInt()-1;
            int r = sc.nextInt()-1;
            int cntOnes = cumOnes[r];
            int cntZeros = cumZeros[r];
            if(l > 0){
                cntOnes -= cumOnes[l-1];
                cntZeros -= cumZeros[l-1];
            }

            int ans = adj(pow2[cntOnes]-1);

            ans = adj(ans + adj(1l*ans*(adj(pow2[cntZeros]-1))));
            pw.println(ans);
        }


        pw.flush();
        pw.close();
    }

    static int adj(long a){
        return (int)(((a%mod) + mod) %mod);
    }
    static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            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;
        }
    }
}