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.util.*;

public class Main {
    private static final int MOD = 1000000007;
    private static int add(long a, long b) {
        return (int) ((a%MOD+b%MOD)%MOD);
    }
    private static int subtract(long a, long b) {
        return add(a, -b);
    }
    private static int multiply(long a, long b) {
        return (int) ((a%MOD*b%MOD)%MOD);
    }
    private static int power(long a, long b) {
        long c = 1;
        while(b > 0) {
            if((b&1) > 0) {
                c = multiply(c, a);
            }
            a = multiply(a, a);
            b >>= 1;
        }
        return (int) c;
    }
    public static void main(String[] args) throws IOException{
        //Scanner f = new Scanner(new File("uva.in"));
        //Scanner f = new Scanner(System.in);
        //BufferedReader f = new BufferedReader(new FileReader("uva.in"));
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        StringTokenizer st = new StringTokenizer(f.readLine());
        int n = Integer.parseInt(st.nextToken());
        int q = Integer.parseInt(st.nextToken());
        int[] prefixSum = new int[n+1];
        char[] s = f.readLine().toCharArray();
        for(int i = 1; i <= n; i++) {
            prefixSum[i] = prefixSum[i-1]+(s[i-1] == '1' ? 1 : 0);
        }
        for(int i = 0; i < q; i++) {
            st = new StringTokenizer(f.readLine());
            int l = Integer.parseInt(st.nextToken());
            int r = Integer.parseInt(st.nextToken());
            out.println(multiply(subtract(power(2, prefixSum[r]-prefixSum[l-1]), 1), power(2, r-l-prefixSum[r]+prefixSum[l-1]+1)));
        }
        f.close();
        out.close();
    }
}