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.util.*;

public class Main {
    int [] pre;
    void solve() throws IOException {
        int n =readInt();
        int q =readInt();
        long mod = 1000_000_007;
        long pow2[] = new long[1000_000];
        pow2[0]=1;
        for(int i =1;i<1000_000;i++){
            pow2[i]=2*pow2[i-1];
            pow2[i]%=mod;
        }
        char [] s =readString().toCharArray();
        pre = new int [n];
        for(int i =0;i<n;i++){
            if(s[i]=='1') pre[i]++;
            if(i>0){
                pre[i]+=pre[i-1];
            }
        }
        int l,r;
        while(q-->0)
        {
            l=readInt()-1;
            r=readInt()-1;
            int len = r-l+1;
            int units = get(r,l);
            long req = pow2[units]-1;
            long zero =  pow2[units]-1;
            int z = len - units;
            zero*=pow2[z]-1;
            zero%=mod;
            out.println((req+zero)%mod);
        }
    }
    int get(int r, int l){
        if(l>0) return pre[r]-pre[l-1];
        return pre[r];
    }

    void init() throws FileNotFoundException {
        try {
            in = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter("output.txt");
        } catch (Exception e) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        }
        tok = new
                StringTokenizer("");
    }

    int minInt(int... values) {
        int min = Integer.MAX_VALUE;
        for (int value : values) {
            min = Math.min(min, value);
        }
        return min;
    }

    int maxInt(int... values) {
        int max = Integer.MIN_VALUE;
        for (int value : values) {
            max = Math.max(max, value);
        }
        return max;
    }

    long minLong(long... values) {
        long min = Long.MAX_VALUE;
        for (long value : values) {
            min = Math.min(min, value);
        }
        return min;
    }

    long maxLong(long... values) {
        long max = Long.MIN_VALUE;
        for (long value : values) {
            max = Math.max(max, value);
        }
        return max;
    }

    boolean checkIndex(int index, int size) {
        return (0 <= index && index < size);
    }

    int abs(int a) {
        if (a < 0) return -a;
        return a;
    }

    long abs(long a) {
        if (a < 0) return -a;
        return a;
    }

    int gcd(int a, int b) {
        int tmp = 0;
        while (b != 0) {
            tmp = a;
            a = b;
            b = tmp % a;
        }
        return a;
    }

    long gcd(long a, long b) {
        long tmp = 0;
        while (b != 0) {
            tmp = a;
            a = b;
            b = tmp % a;
        }
        return a;
    }

    public static void main(String[] args) {
        new Main().run();
    }

    BufferedReader in;
    PrintWriter out;
    StringTokenizer tok;


    void run() {
        try {
            long timeStart = System.currentTimeMillis();
            init();
            solve();
            out.close();
            long timeEnd = System.currentTimeMillis();
            System.err.println("time = " + (timeEnd - timeStart) + " compiled");
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(-1);
        }


    }

    String readLine() throws IOException {
        return in.readLine();
    }

    String delimiter = " ";

    String readString() throws IOException {
        while (!tok.hasMoreTokens()) {
            String nextLine = readLine();
            if (null == nextLine)
                return null;

            tok = new StringTokenizer(nextLine);
        }

        return tok.nextToken(delimiter);
    }

    int[] readIntArray(int b) {
        int a[] = new int[b];
        for (int i = 0; i < b; i++) {
            try {
                a[i] = readInt();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return a;
    }

    double readDouble() throws IOException {
        return Double.parseDouble(readString());
    }

    int readInt() throws IOException {
        return Integer.parseInt(readString());
    }

    long readLong() throws IOException {
        return Long.parseLong(readString());
    }

}