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

// Solution to PCS Problem 89: Banh Mi
// https://pcs.cs.cloud.vt.edu/contests/89/problems/C
// Solution by: jLink23

public class BanhMi {
  private static PrintWriter out;
  public static final int MOD = 1000000007;

  public static void main(String[] args) {
    out = new PrintWriter(System.out);
    FastScanner in = new FastScanner();

    int n = in.nextInt();
    int q = in.nextInt();
    String banhmi = in.next();

    // Create prefix sum array
    int[] prefix = new int[n + 1];
    for (int i = 1; i <= n; i++) {
      if (banhmi.charAt(i-1) == '0') {
        prefix[i] = prefix[i-1];
      } else {
        prefix[i] = prefix[i-1] + 1;
      }
    }

    Long[] pow2 = new Long[100001];
    pow2[0] = 1L;
    for (int i = 1; i <= 100000; i++) {
      pow2[i] = pow2[i-1] * 2L % MOD;
    }

    for (int i = 0; i < q; i++) {
      int l = in.nextInt();
      int r = in.nextInt();
      int k = r-l+1;

      int x = prefix[r] - prefix[l-1];
      int y = k - x;

      out.println(((pow2[x] - 1) * pow2[y]) % MOD);

      // BigInteger b1 = new BigInteger("2");
      // BigInteger b2 = new BigInteger("1");
      // out.println((b1.pow(x).subtract(b2).multiply(b1.pow(y)))
      //   .mod(BigInteger.valueOf((long) Math.pow(10,9) + 7)));
    }
    // out.println(Arrays.toString(prefix));

    out.flush();
  }

  public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(Reader in) {
            br = new BufferedReader(in);
        }

        public FastScanner() {
            this(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 readNextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        int[] readIntArray(int n) {
            int[] a = new int[n];
            for (int idx = 0; idx < n; idx++) {
                a[idx] = nextInt();
            }
            return a;
        }
    }

}