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.*;
                                 
public class Main {
                             
    static Scanner in;
         
//  static StreamTokenizer in;
//  static int nextInt() throws IOException {
//      in.nextToken();
//      return (int)in.nval;    
//  }
    static Random rand = new Random();
    static long gcd(long a, long c) {
        BigInteger q = BigInteger.valueOf(a);
        BigInteger w = BigInteger.valueOf(c);
        q = q.gcd(w);
        return q.longValue();
    }
    static int[] w, d;
   
    public static void main(String args[]) throws IOException {
    	
//      in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        in = new Scanner(System.in);         
        PrintWriter out = new PrintWriter(System.out);
        int n = in.nextInt();
        int q = in.nextInt();
        String s = in.next();
        int[] a = new int[n];
        long[] pow = new long[n];
        pow[0] = 1;
       	long two = 1;
        int mod = (int) (1e9+7);
        for(int i = 0;i<n;i++) {
        	a[i] = s.charAt(i) -'0';
        }
        for(int i=1;i<n;i++) {
        	a[i] = a[i]+a[i-1];
        	two*=2;
        	two%=mod;
        	pow[i] = (pow[i-1] + two)%mod;
        }
        for(int h=0;h<q;h++) {
        	int l = in.nextInt()-1;
        	int r = in.nextInt()-1;
        	int k =0;
        	if(l==0) k = a[r];
        	else k = a[r] - a[l-1];
        	long ans = 0;
        	if(r - k - l == -1) ans = pow[r-l];
        	else ans = (pow[r-l] - pow[r-k-l] +mod) % mod;
        	out.println(ans);
        }
        out.close();
    }
}
                                 
class Pair implements Comparable{
	int a;
    int b;

    public Pair() {
        this.a = 0;
        this.b = 0;
    }
    public Pair(int a, int b) {
        this.a = a;
        this.b = b;

    }
    public int compareTo(Object arg0) {
        if(Integer.compare(this.a, ((Pair) arg0).a) == 0) return Integer.compare(this.b, ((Pair) arg0).b);
            return Integer.compare(this.a, ((Pair) arg0).a);
    }
}