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 Main
{

    public static void process()throws IOException
    {
        int n=ni(),q=ni();
        char arr[]=(" "+nln()).toCharArray();
        int pref_0[]=new int[n+2],pref_1[]=new int[n+2];

        for(int i=1;i<=n;i++){
            if(arr[i]=='0')
                pref_0[i]++;
            else
                pref_1[i]++;
            pref_0[i]+=pref_0[i-1];
            pref_1[i]+=pref_1[i-1];
        }

        while(q-- > 0){
            int l=ni(),r=ni();

            int one=pref_1[r]-pref_1[l-1],zero=pref_0[r]-pref_0[l-1];

            long res=0l,sum=0l;

            sum=(long)mpow(2, one)-1l;
            sum=(sum+mod)%mod;

            res=sum;

            res+=((sum*((mpow(2,zero)-1l+mod)))%mod)%mod;
            res=res%mod;
            pn(res);
        }
    }


    static AnotherReader sc;
    static PrintWriter out;
    public static void main(String[]args)throws IOException
    {
        out = new PrintWriter(System.out);
        sc=new AnotherReader();
        boolean oj = true;

        oj = System.getProperty("ONLINE_JUDGE") != null;
        if(!oj) sc=new AnotherReader(100);

        long s = System.currentTimeMillis();
        int t=1;
       // t=ni();
        while(t-->0)
            process();
        out.flush();
        if(!oj)
            System.out.println(System.currentTimeMillis()-s+"ms");
        System.out.close();  
    }
    
    
    static void pn(Object o){out.println(o);}
    static void p(Object o){out.print(o);}
    static void pni(Object o){out.println(o);System.out.flush();}
    static int ni()throws IOException{return sc.nextInt();}
    static long nl()throws IOException{return sc.nextLong();}
    static double nd()throws IOException{return sc.nextDouble();}
    static String nln()throws IOException{return sc.nextLine();}
    static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int bit(long n)throws IOException{return (n==0)?0:(1+bit(n&(n-1)));}
    static boolean multipleTC=false;
    static long mod=(long)1e9+7l;

    static void r_sort(int arr[],int n){
        Random r = new Random(); 
        for (int i = n-1; i > 0; i--){ 
            int j = r.nextInt(i+1); 
                  
            int temp = arr[i]; 
            arr[i] = arr[j]; 
            arr[j] = temp; 
        } 
        Arrays.sort(arr); 
    }

    static long mpow(long x, long n) {
        if(n == 0)
            return 1;
        if(n % 2 == 0) {
            long root = mpow(x, n / 2);
            return root * root % mod;
        }else {
            return x * mpow(x, n - 1) % mod;
        }
    }
    
    static long mcomb(long a, long b) {
        if(b > a - b)
            return mcomb(a, a - b);
        long m = 1;
        long d = 1;
        long i;
        for(i = 0; i < b; i++) {
            m *= (a - i);
            m %= mod;
            d *= (i + 1);
            d %= mod;
        }
        long ans = m * mpow(d, mod - 2) % mod;
        return ans;
    }
/////////////////////////////////////////////////////////////////////////////////////////////////////////

    static class AnotherReader{BufferedReader br; StringTokenizer st;
    AnotherReader()throws FileNotFoundException{
    br=new BufferedReader(new InputStreamReader(System.in));}
    AnotherReader(int a)throws FileNotFoundException{
    br = new BufferedReader(new FileReader("input.txt"));}
    String next()throws IOException{
    while (st == null || !st.hasMoreElements()) {try{
    st = new StringTokenizer(br.readLine());}
    catch (IOException  e){ e.printStackTrace(); }}
    return st.nextToken(); } int nextInt() throws IOException{
    return Integer.parseInt(next());}
    long nextLong() throws IOException
    {return Long.parseLong(next());}
    double nextDouble()throws IOException { return Double.parseDouble(next()); }
    String nextLine() throws IOException{ String str = ""; try{
    str = br.readLine();} catch (IOException e){
    e.printStackTrace();} return str;}}
/////////////////////////////////////////////////////////////////////////////////////////////////////////////
}