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 {
    static int n, q;
    static boolean[] arr;
    static query[] qs;
    static long[] out, pows;
    static long mod = (long)(1e9 + 7), modi = 500000004L, ans;
	public static void main (String[] args) throws IOException {
	    //pre comp pows of two mod
	    pows = new long[100_000];
	    pows[0] = 1;
	    for(int i = 1; i < 100_000; ++i){
	        pows[i] = (pows[i - 1] * 2) % mod;
	    }
	    
	    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	    PrintWriter pw = new PrintWriter(System.out);
	    
	    String[] nq = br.readLine().split(" ");
	    n = p(nq[0]);
	    q = p(nq[1]);
	    
	    arr = new boolean[n];
	    qs = new query[q];
	    out = new long[q];
	    
	    char[] in = br.readLine().toCharArray();
	    for(int i = 0; i < n; ++i) arr[i] = in[i] == '1';
	    
	    for(int i = 0; i < q; ++i){
	        String[] curr = br.readLine().split(" ");
	        int ll = p(curr[0]) - 1, rr = p(curr[1]) - 1;
	        qs[i] = new query(ll, rr, i);
	    }
	    Arrays.sort(qs);
	    
	    int cl = 0, cr = 0;
	    ans = (arr[0] ? 1 : 0);
	    
	    for(int i = 0; i < q; ++i){
	        int l = qs[i].l, r = qs[i].r;
	        
	        while(cl > l){
	            --cl;
	            if(!arr[cl]) mul2();
	            else add(cr - cl + 1);
	        }
	        
	        while(cr < r){
	            ++cr;
	            if(!arr[cr]) mul2();
	            else add(cr - cl + 1);
	        }
	        
	        while(cl < l){
	            if(!arr[cl]) div2();
	            else sub(cr - cl + 1);
	            ++cl;
	        }
	        
	        while(cr > r){
	            if(!arr[cr]) div2();
	            else sub(cr - cl + 1);
	            --cr;
	        }
	        
	        out[qs[i].id] = ans;
	    }
	    
	    for(int i = 0; i < q; ++i) pw.println(out[i]);
	    pw.close();
	}
	static void add(int size){
	    ans = (ans + pows[size - 1]) % mod;
	}
	static void sub(int size){
	    ans -= pows[size - 1];
	    if(ans < 0) ans += mod;
	}
	static void mul2(){
	    ans = (ans * 2) % mod;
	}
	static void div2(){
	    ans = (ans * modi) % mod;
	}
	static int p(String in){
	    return Integer.parseInt(in);
	}
	static class query implements Comparable<query> {
	    int l, r, id;
	    query(int ll, int rr, int ii){
	        l = ll;
	        r = rr;
	        id = ii;
	    }
	    public int compareTo(query in){
	        int lbuck1 = (int)(l / Math.sqrt(n));
	        int lbuck2 = (int)(in.l / Math.sqrt(n));
	        if(lbuck1 != lbuck2) return lbuck1 - lbuck2;
	        
	        if(lbuck1 % 2 == 0) return r - in.r;
	        return in.r - r;
	    }
	}
}