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.
import java.util.*;
import java.awt.geom.*;
import java.io.*;
import java.math.*;

public class Main
{

  public static void main(String[] args) throws Exception
  {
    long startTime = System.nanoTime();
    int n = in.nextInt();
    int q = in.nextInt();
    int [] zeroes = new int [n];
    int [] ones = new int [n];
    long MOD = 1_000_000_000 + 7;
    char [] data = in.next().toCharArray();
    if(data[0]=='0') zeroes[0] = 1;
    if(data[0]=='1') ones[0] = 1;
    for(int i=1;i<data.length;i++){
        if(data[i]=='0'){
            zeroes[i]  =  zeroes[i-1]  + 1;
            ones[i]  =  ones[i-1] ;
        }
        else{
            zeroes[i]  =  zeroes[i-1] ;
            ones[i]  =  ones[i-1]  + 1;
        }        
    }
    for(int i=0;i<q;i++){
        int l = in.nextInt();
        int r = in.nextInt();
        long z = (zeroes[r-1]-zeroes[l-1]) + (data[l-1]=='0' ? 1 : 0);
        long o  = (ones[r-1]-ones[l-1]) + (data[l-1]=='1' ? 1 : 0);
        long temp = mod_pow(2L,z,MOD);
        temp%=MOD;
        long ans = (mod_pow(2L,o,MOD) - 1)%MOD * temp;
        ans%=MOD;
        out.println(ans);
    }
    long endTime = System.nanoTime();
    err.println("Execution Time : +" + (endTime-startTime)/1000000 + " ms");
    exit(0);
  }

  static long mod_pow(long base,long exponent,long mod)
    {
      if(mod==1) return 0;
      long result=1;
      base%=mod;
      while(exponent>0)
      {
        if(exponent%2==1)
        {
          result=(result*base)%mod;
        }
        exponent=(exponent>>1);
        base=(base*base)%mod;
      }
      return result;
    }


  static class InputReader
  {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

        public double nextDouble(){
           return Double.parseDouble(next());
        }
    }

    static void exit(int a)
    {
      out.close();
      err.close();
      System.exit(a);
    }

    static InputStream inputStream = System.in;
    static OutputStream outputStream = System.out;
    static OutputStream errStream = System.err;
    static InputReader in = new InputReader(inputStream);
    static PrintWriter out = new PrintWriter(outputStream);
    static PrintWriter err = new PrintWriter(errStream);
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

}
