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.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;

public class Main {

	private static class Solution {
		int[] arr;
		int[] pre;
		int n;

		int[] p2;

		int mod = 1_000_000_000 + 7;

		public Solution(int[] arr) {
			this.arr = arr;
			n = arr.length - 1;
			pre = new int[n + 1];
			for (int i = 1; i <= n; ++i) {
				pre[i] = pre[i - 1] + arr[i];
			}

			p2 = new int[n + 1];
			p2[0] = 1;
			for (int i = 1; i <= n; ++i) {
				p2[i] = (p2[i - 1] * 2) % mod;
			}

		}

		public int query(int low, int high) {

			int cnt1 = pre[high] - pre[low - 1];
			if (cnt1 == 0)
				return 0;

			int cnt0 = high - low + 1 - cnt1;

			long ret = (long) ((p2[cnt1] - 1)) * p2[cnt0];
			ret %= mod;
			if (ret < 0)
				ret += mod;
			return (int) ret;
		}
	}

	public static void main(String[] args) throws Exception {
		MyScanner in = new MyScanner();

		int n = in.nextInt();
		int q = in.nextInt();

		String str = in.next();
		int[] arr = new int[n + 1];
		for (int i = 1; i <= n; ++i) {
			arr[i] = str.charAt(i - 1) - '0';
		}

		Solution sol = new Solution(arr);

		StringBuilder sb = new StringBuilder();
		while (q-- > 0) {
			int low = in.nextInt();
			int high = in.nextInt();

			sb.append(sol.query(low, high) + "\n");
		}

		System.out.print(sb.toString());
	}

	// -----------MyScanner class for faster input----------
	public 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;
		}

	}
	// --------------------------------------------------------
}