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:
//package math_codet;

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

public class lets_do {
    InputReader in;
    PrintWriter out;
    Helper_class h;
    final long mod = 1000000007;
    public static void main(String[] args) throws java.lang.Exception{
        new lets_do().run();
    }
    void run() throws Exception{
        in=new InputReader();
        out = new PrintWriter(System.out);
        h = new Helper_class();
        int t = 1;
        while(t-->0)
            solve();
        out.flush();
        out.close();
    }
    void solve(){
        int n = h.ni();
        int q = h.ni();
        int i = 0;
        String st = h.n();
        int[] pf = new int[n];
        pf[0] = st.charAt(0) - '0';
        for(i = 1; i < n; i++)
            pf[i] = pf[i - 1] + st.charAt(i) - '0';
        while(q-->0){
            int l = h.ni() - 1;
            int r = h.ni() - 1;
            int ones = l == 0 ? pf[r] : pf[r] - pf[l - 1];
            int zeroes = r - l + 1 - ones;
            //h.pn(ones+" "+zeroes);
            long xx = (h.modPow(2, ones) - 1 + mod) % mod;
            long yy = (h.modPow(2, zeroes) - 1 + mod) % mod;
            xx = (xx * (yy + 1)) % mod;
            h.pn(xx);
        }
    }
    static final Comparator<Entity> com=new Comparator<Entity>(){
        public int compare(Entity a, Entity b){
            int xx=Integer.compare(a.x,b.x);
            if(xx==0)
                return Integer.compare(a.y,b.y);
            else
                return xx;
        }
    };
    class Entity{
        int x, y;
        Entity(int p, int q){
            x=p;
            y=q;
        }
    }
    class Helper_class{
        long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
        int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
        int bitcount(long n){return (n==0)?0:(1+bitcount(n&(n-1)));}
        void p(Object o){out.print(o);}
        void pn(Object o){out.println(o);}
        void pni(Object o){out.println(o);out.flush();}
        String n(){return in.next();}
        String nln(){return in.nextLine();}
        int ni(){return Integer.parseInt(in.next());}
        long nl(){return Long.parseLong(in.next());}
        double nd(){return Double.parseDouble(in.next());}

        long mul(long a,long b){
            if(a>=mod)a%=mod;
            if(b>=mod)b%=mod;
            a*=b;
            if(a>=mod)a%=mod;
            return a;
        }
        long modPow(long a, long p){
            long o = 1;
            while(p>0){
                if((p&1)==1)o = mul(o,a);
                a = mul(a,a);
                p>>=1;
            }
            return o;
        }
        long add(long a, long b){
            if(a>=mod)a%=mod;
            if(b>=mod)b%=mod;
            if(b<0)b+=mod;
            a+=b;
            if(a>=mod)a-=mod;
            return a;
        }
    }

    class InputReader{
        BufferedReader br;
        StringTokenizer st;
        public InputReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public InputReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        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;
        }
    }
}