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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.Scanner;
import java.util.StringTokenizer;

public class main {
	 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; 
	        } 
	    } 
static long mod = 1000000007;
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastReader in = new FastReader();
		int n = in.nextInt();
		int q = in.nextInt();
		//in.nextLine();
		long num[] = new long[100001];
		int c= 2;
		num[0]=1;	
		for(int i = 1;i<100000;i++) {
			num[i] = 2*((num[i-1])%mod)%mod;
		}
		String s = in.nextLine();
		int[] nums = new int[s.length()+1];
		int[] numsz = new int[s.length()+1];
		//nums[0]=s.charAt(0)=='1'?1:0;
		for(int i = 1;i<s.length()+1;i++) {
			if(s.charAt(i-1)=='1')
			nums[i]=nums[i-1]+1;
			else
				nums[i]=nums[i-1];
			if(s.charAt(i-1)=='0')
				numsz[i]=numsz[i-1]+1;
			else
				numsz[i]=numsz[i-1];
		}
		int t = 0;
		while(t<q) {
			int in1 = in.nextInt();
			int in2 = in.nextInt();
			int ones = nums[in2]-nums[in1-1];
			int z = in2-in1+1-(ones);
			long p = (fastpow(2L,ones)-1L)%mod*(fastpow(2L,z)%mod)%mod;
//			BigInteger c = new BigInteger("2");
//			BigInteger d = new BigInteger("2");
//			c = c.pow(ones+1).subtract(new BigInteger("1")).multiply(d.pow(in2-in1+1-(ones+1))).mod(new BigInteger("1000000007"));;
			//c= c.mod(new BigInteger("1000000007"));//(Math.pow(2, in2-in1+1)%1000000007)-(Math.pow(2, in2-in1+1-ones)%1000000007);
			System.out.println(p);
			t++;
		}
	}
	static long fastpow(long base, long pow)
	{
	    if (pow == 0)
	        return 1;
	    if (base == 0)
	        return 0;
	    if (pow == 1)
	        return base;
	    long x = fastpow(base, pow >>1) % mod;
	    if (pow % 2 == 0)
	        return ((x % mod) *(x % mod) % mod);
	    else
	        return ((((x % mod)* (x % mod) % mod) % mod) * (base % mod) % mod);
	}
}