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 yuy {

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int n=sc.nextInt();
		int q=sc.nextInt();
		String h=sc.nextLine();
		int[] pow2 = new int[1000000];
        pow2[0] = 1;
        for (int i = 1; i < pow2.length; i++) {
            pow2[i] = (int)((1l*pow2[i-1]*2)%mod);
        }
		count [] a=new count[n]; int o=0; int z=0;
		for(int i=0;i<n;i++) {
			if(h.charAt(i)=='1')
				o++;
			else if(h.charAt(i)=='0')
					z++;
			a[i]=new count(o,z);
		}
	    long ans=1;
	 //   System.out.println(Arrays.toString(a));
	    for(int i=0;i<q;i++) {
	    	int l=sc.nextInt()-1;
	        int r=sc.nextInt()-1;
	        long x=0;
	        long y=0;
	        l--;
	        ans=1;
	        if(l==-1) {
	        	 x= a[r].one; y=a[r].zero;
	        	 /*while(x>0) {long w=x;
        		 if(x>60)
        			 w=60;
	        		 ans=((long)Math.pow(2,w)*ans)%1000000009;
	        		 x=x-60;
	        	 }*/
	        	 ans=(pow2[(int)x]-1l)*pow2[(int)y]%mod;
	        }
	        else {
	        	x= a[r].one-a[l].one; y=a[r].zero-a[l].zero;
	        	ans=(pow2[(int)x]-1l)*pow2[(int)y]%mod;
	        }
	        out.println(ans%mod);
	    }
	    
		out.flush();
	}
	static int mod = (int)1e9 + 7;
	static class count{
		int one;
		int zero;
		count(int o,int z){
			one=o;
			zero=z;
		}
		public String toString() {
			return one+" "+zero;
		}
	}
	static 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 String nextLine() throws IOException {
			return br.readLine();
		}

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

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

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

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

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}