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.util.*;
import java.io.*;



public class Main
{

	public static int gcd(int a , int b)
	{
		int divs=0;
		int divd=0;
		if(a>=b)
		{
			divd=a;
			divs=b;
		}
		else
		{
			divd=b;
			divs=a;
		}

		int rem=divd%divs;
		while(rem!=0)
		{
			int temp=divs;
			divs=rem;
			divd=temp;
			rem=divd%divs;
		}
		return divs;

	}
	static class spComp implements Comparator<long[]>
	{
		public int compare(long[] a, long[] b)
		{
			
			if(a[0]>b[0])
				return 1;
			else if(a[0]<b[0])
				return -1;
			else if(a[1]>b[1])
				return 1;
			else if(a[1]>b[1])
				return -1;
			else return 1;
		}
	}

	class unionFind
	{

		int parent[];
		int height[];
		int size=0;
		unionFind(int n)
		{
			size=n;
			parent = new int[n];
			height = new int[n];
			// Arrays.fill(parent, -1);
			for(int i=0;i<n;i++)
				parent[i]=i;

		}
		int find(int a)
		{
			if(parent[a]==a)
				return a;

			parent[a]=find(a);

			return parent[a];
		}

		void add(int a, int b)
		{
			a = find(a);
			b = find(b);
			
			if(height[a] > height[b])
			{
				parent[b] = a;
			}
			else if(height[ b ] > height[ a ])
				parent[ a ] = b;
			else if(height[ a ] == height[ b ])
			{
				height[ a ] = height[ a ]+1;
				parent[ b ] = a;
			}


		}
	}


	static ArrayList<Integer> largestpow(int val, ArrayList<Integer> primes)
	{
		int n=primes.size();
		ArrayList<Integer> powers= new ArrayList<Integer>();

		for(int i=0;i<n;i++)
		{
			int p=primes.get(i);
			while(val%p==0)
			{
				p=p*p;
			}
			powers.add(p);
		}
		return powers;
	}


	public static int twoPow(int a)
	{
		
		int val=1;
		int ans=0;
		if(a==1)
			return 1;
		while(val<a)
		{
			val*=2;
			ans++;
		}
		if(val==a)
			return ans;
		else return ans;

		// return ans;
	}


	static long pow2(int a)
	{
		long modd=1000000007;
		long ans=1;
		for(int i=0;i<a;i++)
		{
			ans*=2;
			ans%=modd;
		}
		
		return ans;
	}

	public static void main(String args[])
	{
		Scanner sc= new Scanner(System.in);
		int n= sc.nextInt();
		int q= sc.nextInt();
		int arr[]= new int[n];
		String str= sc.next();
		for(int i=0;i<n;i++)
		{
			arr[i]= str.charAt(i)-'0';
		}
		int zeros[]= new int[n];
		int ones[]= new int[n];
		long modd= 1000000007;
		long powers[]= new long[n+1];
		for(int i=0;i<=n;i++)
		{
			if(i==0)
				powers[i]=1;
			else
			powers[i]= (powers[i-1]*2)%modd;
		}
		for(int i=0;i<n;i++)
		{
			if(i==0)
			{
				if(arr[i]==0)
					zeros[i]=1;
				if(arr[i]==1)
					ones[i]=1;
			}
			else
			{
				if(arr[i]==0)
				{
					zeros[i]=zeros[i-1]+1;
					ones[i]=ones[i-1];
				}
				if(arr[i]==1)
				{
					ones[i]= ones[i-1]+1;
					zeros[i]=zeros[i-1];
				}
			}
		}
		
		for(int i=0;i<q;i++)
		{
			int l= sc.nextInt()-1;
			int r= sc.nextInt()-1;
			if(l!=0)
			{
				int zer=zeros[r]-zeros[l-1];
				int one= ones[r]-ones[l-1];
				long ans= (powers[one+zer]-powers[zer] + modd)%modd;
				System.out.println(ans);

			}
			else
			{
				int zer=zeros[r];
				int one= ones[r];
				long ans= (powers[one+zer]-powers[zer] + modd)%modd;
				System.out.println(ans);
			}
		}
		
	}



}