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:
//package graphs;
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class PW {
	
	static class FastReader {
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            br = new BufferedReader(
                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 nextLine()
        {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastReader s = new FastReader();
		
		//int a=s.nextInt();
		//lon n=s.nextLong();
		
		int n=s.nextInt();
		int q=s.nextInt();
		long mod=1000000007;
		
		String str=s.next();
		long a[]=new long[str.length()];
		long ans=0;
		for(int i=0;i<a.length;i++)
		{
			if(str.charAt(i)=='1')
			{
				if(i==0)
					a[i]=1;
				else
					a[i]=1+a[i-1];
			}
			else
			{
				if(i-1>=0)
				a[i]=a[i-1];
			}
				
		}
		
		for(int i=0;i<q;i++)
		{
			int l=s.nextInt();
			int r=s.nextInt();
			ans=0;
			l--;r--;
			
			long ones=0;
			if(l==0)
				ones=a[r];
			else
			 ones=a[r]-a[l-1];
			
			long zeros=(r-l+1)-ones;
			//System.out.println("z "+zeros+"o "+ones);
			ans=(ans%mod+(( (fastExpo(2,ones,mod)-1)%mod) * (fastExpo(2,zeros,mod))%mod)%mod)%mod;
			System.out.println(ans);
			
		}
		
		
		
	}
	
	
	
	public static long solve(int n, int r)
    {


        long p = 1, k = 1;
        if (n - r < r) {
            r = n - r;
        }
 
        if (r != 0) {
            while (r > 0) {
                p *= n;
                k *= r;
 
                long m = __gcd(p, k);
 
                
                p /= m;
                k /= m;
 
                n--;
                r--;
            }
 
           
        }
        else {
            p = 1;
        }
 
        
        //System.out.println(p);
        return p;
    }
	
	
	public static long gcd(long a,long b)
	{
		if(a==0||b==0)
			return a+b;
		return gcd(b,(a%b));
	}
	
	public static long __gcd(long n1, long n2)
    {
        long gcd = 1;
 
        for (int i = 1; i <= n1 && i <= n2; ++i) {
            // Checks if i is factor of both integers
            if (n1 % i == 0 && n2 % i == 0) {
                gcd = i;
            }
        }
        return gcd;
    }
	
	public static boolean prime(int n)
	{
		if(n<=2)
			return true;
		
		for(int i=2;i<=Math.sqrt(n);i++)
		{
			if(n%i==0)
				return false;
		}
		
		return true;
		
	}
	public static long fastExpo(long a,long n,long mod){
        if (n == 0)
            return 1;
        else{
            long x = fastExpo(a,n/2,mod);
            if ((n&1) == 1){
                return (((a*x)%mod)*x)%mod;
            }
            else{
                return (((x%mod)*(x%mod))%mod)%mod;
            }
        }
    }
}
class pair{
	//public:
		long f;
		long s;
		pair(long x,long y)
		{
			f=x;
			s=y;
		}
        
}