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.*;
import java.math.*;

public class C{

    static void precompute(int n){
        pow2 = new long[n + 1];
        pow2[0] = 1l;
        for(int i = 1; i <= n; i++)
            pow2[i] = (pow2[i - 1] * 2l) % mod;
    }

	public static void process(int testNumber){
        int n = ni(), q = ni();
        precompute(n);
        char arr[] = (" " + nln()).toCharArray();
        long pref[] = new long[n + 1];
        for(int i = 1; i <= n; i++){
            pref[i] += pref[i - 1] + (arr[i] - '0');
        }	

        for(int query = 1; query <= q; query++){
            int l = ni(), r = ni();
            long ones = pref[r] - pref[l - 1], zeroes = (r - l + 1)*1l - ones, res = 0l;
            res += pow2[(int)ones] - 1;
            res += mod;
            res %= mod;
            long x = res;
            res += (x * ((pow2[(int)zeroes] - 1) + mod)%mod) % mod;
            res %= mod;
            pn(res);
        }
	}
    
    static long pow2[];

	static final long mod = (long)1e9+7l;
	static boolean DEBUG = true;
	static FastReader sc;
	static PrintWriter out = new PrintWriter(System.out);
	public static void main(String[]args){
		sc = new FastReader();
 
		long s = System.currentTimeMillis();
		int t = 1;
		// t = ni();
		for(int i = 1; i <= t; i++)
			process(i);
 
		out.flush();
		System.err.println(System.currentTimeMillis()-s+"ms");
	}

	static void trace(Object... o){ if(!DEBUG) return; System.err.println(Arrays.deepToString(o)); };    
	static void pn(Object o){ out.println(o); }
	static void p(Object o){ out.print(o); }
	static int ni(){ return Integer.parseInt(sc.next()); }
	static long nl(){ return Long.parseLong(sc.next()); }
	static double nd(){ return Double.parseDouble(sc.next()); }
	static String nln(){ return sc.nextLine(); }
	static long gcd(long a, long b){ return (b==0)?a:gcd(b,a%b);}
	static int gcd(int a, int b){ return (b==0)?a:gcd(b,a%b); }
	
	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(); 
		} 
  
		String nextLine(){ 
			String str = ""; 
			try{ str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } 
			return str; 
		} 
	} 
}

class pair implements Comparable<pair> {
    int first, second;
    public pair(int first, int second){
        this.first = first;
        this.second = second;
    }

    @Override
    public int compareTo(pair ob){
        if(this.first != ob.first)
            return this.first - ob.first;
        return this.second - ob.second;
    }

    static public pair from(int f, int s){
        return new pair(f, s);
    }
}