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.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;


public class Solution {
static  long mod = (int)1e9 + 7 ;






    public static void main(String[] args) {


        int n = fsca.nextInt();
        int q = fsca.nextInt();
        char[] s = fsca.next().toCharArray();
        int[] sum = new int[s.length + 1];
        long[] e2 = new long[s.length + 1];
        long mod = 1000000007L;
        e2[0] = 1;
        for (int i = 1; i <= n; ++i) {
            sum[i] = sum[i - 1] + ((s[i - 1] == '1') ? 1 : 0);
            e2[i] = e2[i - 1] * 2;
            e2[i] %= mod;
        }


        while (q-- > 0) {
            int l = fsca.nextInt();
            int r = fsca.nextInt();

            int len = r - l + 1;
            int one = sum[r] - sum[l - 1];
            int zero = len - one;
            long ans = e2[one] - 1;
            long base = ans;
            ans = ans + (e2[zero] - 1) * base % mod;
            ans %= mod;
            fop.println(ans);
        }












            fop.flush();
            fop.close();



    }
    /*-----------------------------------------------------------------------------------------------------------------------------------------------*/

    static PrintWriter fop = new PrintWriter(System.out);
    static  FastScanner fsca = new FastScanner();



    static long gcd(long a, long b) {
        return (b == 0) ? a : gcd(b, a % b);
    }

    ;

    static int gcd(int a, int b) {
        return (b == 0) ? a : gcd(b, a % b);
    }


    static final Random random = new Random();

    static void ruffleSort(int[] a) {
        int n = a.length;//shuffle, then sort
        for (int i = 0; i < n; i++) {
            int oi = random.nextInt(n), temp = a[oi];
            a[oi] = a[i];
            a[i] = temp;
        }
        Arrays.sort(a);
    }

    static void ruffleSort(long[] a) {
        int n = a.length;//shuffle, then sort
        for (int i = 0; i < n; i++) {
            int oi = random.nextInt(n);
            long temp = a[oi];
            a[oi] = a[i];
            a[i] = temp;
        }
        Arrays.sort(a);
    }


    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[][] readMat(int n, int m) {
            int a[][] = new int[n][m];
            for (int i = 0; i < n; i++)
                for (int j = 0; j < m; j++)
                    a[i][j] = nextInt();
            return a;
        }

        char[][] readCharMat(int n, int m) {
            char a[][] = new char[n][m];
            for (int i = 0; i < n; i++) {
                String s = next();
                for (int j = 0; j < m; j++)
                    a[i][j] = s.charAt(j);
            }
            return a;
        }


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

        long[] readLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        long nextLong() {
            return Long.parseLong(next());
        }
    }

    static void print(int a[], int n) {
        for (int i = 0; i < n; i++)
            fop.append((a[i]) + " ");
        // fop.append("\n") ;
    }

    static void print(long a[], int n) {
        for (int i = 0; i < n; i++)
            fop.append((a[i]) + " ");
        // fop.append("\n") ;
    }
}