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:
//package All_in_all;

import java.io.*;
import java.math.BigInteger;
import java.util.*;

/**
 * Created by nikitos on 25.08.17.
 */


public class simvoli {

    public StreamTokenizer t;
    private final int INF = Integer.MAX_VALUE / 2;
    private int mod = 1000000000 + 7;

    public int nextInt() throws IOException {
        t.nextToken();
        return (int) t.nval;
    }

    public long nextLong() throws IOException {
        t.nextToken();
        return (long) t.nval;
    }

    public String nextString() throws IOException {
        t.nextToken();
        return t.sval;
    }

    public void start() throws IOException {
        BufferedReader buf = new BufferedReader(new InputStreamReader(System.in));

        String[] in = buf.readLine().split("[ ]");
        int n = Integer.parseInt(in[0]);
        int q = Integer.parseInt(in[1]);

        String st = buf.readLine();

        int[] pref = new int[n + 1];
        pref[1] = st.charAt(0) == '1'? 1 : 0;
        for (int i = 2; i < n + 1; i++) {
            pref[i] = pref[i - 1] + (st.charAt(i - 1) == '1'? 1 : 0);
        }

        dp = new long[100005];
        fill(dp);

        for (int i = 0; i < q; i++) {
            in = buf.readLine().split("[ ]");
            int from = Integer.parseInt(in[0]);
            int to = Integer.parseInt(in[1]);

            int numOne = pref[to] - pref[from - 1];
            int numAll = to - from + 1;

            if (numOne == 0) {
                System.out.println(0);
                continue;
            }

            long ans = (dp[numAll] - dp[numAll - numOne] + mod) % mod;

            System.out.println(ans);
        }

    }

    private long dp[];
    private int modWas[];

    private void fill(long[] dp) {
        dp[1] = 1;
        for (int i = 2; i < 100005; ++i) {
            dp[i] += 2 * dp[i - 1] + 1;
            
            dp[i] %= mod;
        }
    }

    public static void main(String[] args) throws IOException {
        new simvoli().start();
    }

}