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 A {
	
	public static void main(String[] args) throws Exception{
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int q=sc.nextInt();
		String s=sc.next();
		int arr[]=new int [n];
		for(int i=0;i<n;i++) {
			arr[i]=s.charAt(i)-'0';
		}
		int one[]=new int [n+1];
		int zero[]=new int [n+1];
		if(arr[0]==1)one[1]++;
		else zero[1]++;
		for (int i=1;i<n;i++) {
			one[i+1]+=arr[i]+one[i];
			if(arr[i]==0) {
				zero[i+1]=zero[i]+1;
			}else
				zero[i+1]=zero[i];
		}
		//System.out.println(Arrays.toString(zero));
		while(q-->0) {
			int l=sc.nextInt();
			int r=sc.nextInt();
			int o=one[r]-one[l-1];
			int z=zero[r]-zero[l-1];
		//	System.out.println(o+" "+z);
			long ans=modPow(2,o, (int)1e9+7)-1;
		//	System.out.println(ans);
			long p=1l*modPow(2, o, (int)1e9+7)-1;
		//System.out.println(p+" "+z);
		//	System.out.println((((p%mod)*modPow(2, z, (int)1e9+7))%mod)%mod -7+"**");
			if(z>=1)ans=((((ans%mod)+((p%mod)*(modPow(2, z, (int)1e9+7)-1))%mod)%mod))%mod;
			
			System.out.println((ans+mod)%mod);
		}

	}
	static int mod=(int)1e9+7;
	static long modPow(int a, int e, int mod)	// O(log e)
	{
		a %= mod;
		long res = 1;
		while(e > 0)
		{
			if((e & 1) == 1)
				res =(1l*res * a) % mod;
			a = (int)((1l*a * a) % mod);
			e >>= 1;
		}
		return res;
	}
}
class Scanner{
	StringTokenizer st;
	BufferedReader br;
	public Scanner(InputStream system) { br = new BufferedReader(new InputStreamReader(system));}
	public Scanner(String file) throws Exception{br = new BufferedReader(new FileReader (file));}
	public String next() throws IOException{
		while (st==null|| !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}
	public int nextInt() throws IOException{
		return Integer.parseInt(next());
	}
	public long nextLong() throws IOException{
		return Long.parseLong(next());
	}
	public String nextLine() throws IOException{
		return br.readLine();
	}
	public Double nextDouble() throws IOException{
		return Double.parseDouble(next());
	}

}