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:


/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.StringTokenizer;

/**
 *
 * @author is2ac
 */
public class C_CF {

    public static void main(String[] args) {
        FastScanner57 fs = new FastScanner57();
        PrintWriter pw = new PrintWriter(System.out);
        //int t = fs.ni();
        int t = 1;
        for (int tc = 0; tc < t; tc++) {
            int n = fs.ni(), q = fs.ni();
            long mod = (long)(1e9+7);
            long[] dp = new long[n+1];
            long[] b = new long[n+1];
            dp[0] = 1L;
            b[0] = 1L;
            long p = 1;
            for (int i = 1; i < n+1; i++) {
                p += dp[i-1];
                p %= mod;
                dp[i] = p;
                b[i] = p;
            }
            for (int i = 1; i < n+1; i++) {
                b[i] += b[i-1];
                b[i] %= mod;
            }
            String s = fs.next();
            long[] a = new long[n];
            for (int i= 0; i < s.length(); i++) {
                a[i] += s.charAt(i)-'0';
            }
            for (int i = 1; i < n; i++) {
                a[i] += a[i-1];
            }
            for (int i = 0; i < q; i++) {
                int l = fs.ni()-1, r = fs.ni()-1;
                long o = a[r];
                if (l>0) o -= a[l-1];
                long z = r-l+1 - o;
                if (o==0) {
                    pw.println(0);
                    continue;
                }
                long res = b[(int)o-1];
                if (z>0) {
                    res *= b[(int)z-1];
                } else {
                    res = 0;
                }
                res %= mod;
                res += b[(int)o-1];
                res %= mod;
                pw.println(res);
            }
           //pw.println(Arrays.toString(dp));
            //pw.println(Arrays.toString(b));
        }
        pw.close();

    }

    public static String print(int[] b) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < b.length; i++) {
            sb.append(b[i]);
        }
        return sb.toString();
    }

    public static int recur(int ind, int p, int[] a, Integer[][] dp) {
        if (ind == a.length) {
            return 0;
        }
        int n = (int) (1e9);
        if (dp[ind][p] != null) {
            return dp[ind][p];
        }
        for (int i = 0; i < 3; i++) {
            int c = 1;
            if (i == a[ind]) {
                c--;
            }
            if (i == p) {
                continue;
            }
            n = Math.min(n, recur(ind + 1, i, a, dp) + c);
        }
        return dp[ind][p] = n;
    }

    public static void sort(long[] a) {
        List<Long> list = new ArrayList();
        for (int i = 0; i < a.length; i++) {
            list.add(a[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < a.length; i++) {
            a[i] = list.get(i);
        }
    }

    public static long gcd(long n1, long n2) {
        if (n2 == 0) {
            return n1;
        }
        return gcd(n2, n1 % n2);
    }

}

class UnionFind16 {

    int[] id;

    public UnionFind16(int size) {
        id = new int[size];
        for (int i = 0; i < size; i++) {
            id[i] = i;
        }
    }

    public int find(int p) {
        int root = p;
        while (root != id[root]) {
            root = id[root];
        }
        while (p != root) {
            int next = id[p];
            id[p] = root;
            p = next;
        }
        return root;
    }

    public void union(int p, int q) {
        int a = find(p), b = find(q);
        if (a == b) {
            return;
        }
        id[b] = a;
    }
}

class FastScanner57 {

    BufferedReader br;
    StringTokenizer st;

    public FastScanner57() {
        br = new BufferedReader(new InputStreamReader(System.in), 32768);
        st = null;
    }

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

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

    int[] intArray(int N) {
        int[] ret = new int[N];
        for (int i = 0; i < N; i++) {
            ret[i] = ni();
        }
        return ret;
    }

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

    long[] longArray(int N) {
        long[] ret = new long[N];
        for (int i = 0; i < N; i++) {
            ret[i] = nl();
        }
        return ret;
    }

    double nd() {
        return Double.parseDouble(next());
    }

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}