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

public class CF1062C {
	long exp(long b, long e) {
		if(e == 0) return 1;
		long tmp = exp(b, e / 2);
		tmp = (tmp * tmp) % MOD;
		if((e & 1) == 1) tmp = (tmp * b) % MOD;
		return tmp;
	}
	final int MOD = (int) 1e9 + 7;
	public CF1062C() {
		FS scan = new FS();
		PrintWriter out = new PrintWriter(System.out);
		int n = scan.nextInt(), q = scan.nextInt();
		char[] x = scan.next().toCharArray();
		int[] pre1 = new int[n + 1];
		int[] pre0 = new int[n + 1];
		for(int i = 1 ; i <= n ; i++) {
			pre0[i] = pre0[i - 1] + (x[i - 1] == '0' ? 1 : 0);
			pre1[i] = pre1[i - 1] + (x[i - 1] == '1' ? 1 : 0);
		}
		for(int qq = 0 ; qq < q ; qq++) {
			int l = scan.nextInt(), r = scan.nextInt();
			int z = pre0[r] - pre0[l - 1];
			int o = pre1[r] - pre1[l - 1];
			long res = (exp(2, z) - 1 + MOD) % MOD;
			res *= (exp(2, o) - 1 + MOD) % MOD;
			res %= MOD;
			res += (exp(2, o) - 1 + MOD) % MOD;
			res %= MOD;
			out.println(res);
		}
		out.close();
	}
	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		public String next() {
			while(!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch(Exception e) { e.printStackTrace(); }
			}
			return st.nextToken();
		}
		public int nextInt() { return Integer.parseInt(next()); }
	}
	public static void main(String[] args) { new CF1062C(); }
}