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.BufferedReader;
import java.io.FileReader;
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.Scanner; 
import java.util.StringTokenizer; 
import java.util.*;
import java.io.*;
public class codeforces 
{ 
   static class Student{
       int x,y;
        Student(int x,int y){
            this.x=x;
            this.y=y;
            //this.z=z;
        }
    }
    static int prime[];
    static void sieveOfEratosthenes(int n) 
    { 
        // Create a boolean array "prime[0..n]" and initialize 
        // all entries it as true. A value in prime[i] will 
        // finally be false if i is Not a prime, else true. 
        int pos=0;
        prime= new int[n+1]; 
        for(int p = 2; p*p <=n; p++) 
        { 
            // If prime[p] is not changed, then it is a prime 
            if(prime[p] == 0)
            { 
                // Update all multiples of p
                prime[p]=p;
                for(int i = p*p; i <= n; i += p) 
                    prime[i] = p; 
            } 
        } 
    }
static    class Sortbyroll implements Comparator<Student> 
{ 
    // Used for sorting in ascending order of 
    // roll number 
    public int compare(Student c, Student b) 
    { 
        return c.x-b.x;
    } 
} 
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(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
 
    static class Edge{
        int a,b;
        Edge(int a,int b){
            this.a=a;
            this.b=b;
        }
    }
    static class Trie{
        HashMap<Character,Trie>map;
         int c;
        Trie(){
            map=new HashMap<>();
            //z=null;
            //o=null;
            c=0;
        }
    }
    //static long ans;
    static int parent[];
    static int rank[];
    static int b[][];
    static int bo[];
    static int ho[];
    static int seg[];
    //static int pos;
   // static long mod=1000000007;
    //static int dp[][];
    static HashMap<Integer,Integer>map;
    static PriorityQueue<Student>q=new PriorityQueue<>();
    //static Stack<Integer>st;
   // static ArrayList<Character>ans;
    static ArrayList<ArrayList<Integer>>adj;
    //static long ans;
    static Trie root;
    static long fac[];
    static int gw,gb;
    static long mod=(long)(1000000007);
    //static int ans;
    static void solve()throws IOException{
        FastReader sc=new FastReader();
        int n,q,l,r,o,z,i;
        long ans,t;
        n=sc.nextInt();
        q=sc.nextInt();
        String s=sc.nextLine();
        int a[]=new int[n+1];
        for(i=1;i<=n;i++){
            a[i]=s.charAt(i-1)-'0';
            a[i]+=a[i-1];
        }
        StringBuilder sb=new StringBuilder();
        for(i=0;i<q;i++){
            l=sc.nextInt();
            r=sc.nextInt();
            o=a[r]-a[l-1];
            z=r-l+1-o;
            t=0;
            ans=0;
            t=(power(2,o,mod)-(long)1+mod)%mod;
            ans+=t;
            t=(t*(power(2,z,mod)-(long)1+mod)%mod)%mod;
            ans=(ans+t)%mod;
            sb.append(ans+"\n");
        }
        System.out.println(sb.toString());
    }
    static long nCr(long n, long r, 
                                    long p) 
    { 
        return (fac[(int)n]* modInverse(fac[(int)r], p) 
                % p * modInverse(fac[(int)(n-r)], p) 
                                    % p) % p; 
    } 
    //static int prime[];
    //static int dp[];
    public static void main(String[] args){
        //long sum=0;
        try {
                codeforces.solve();
                } catch (Exception e) {
                    e.printStackTrace();
                }
    }
       static long modInverse(long n, long p) 
    { 
        return power(n, p-(long)2,p); 
    } 
 static long power(long x, long y, long p) 
{ 
    long res = 1;      // Initialize result 
  
    x = x % p;  // Update x if it is more than or  
                // equal to p 
  
    while (y > 0) 
    { 
        // If y is odd, multiply x with result 
        if (y %(long)2!=0) 
            res = (res*x) % p; 
  
        // y must be even now 
        y = y>>1; // y = y/2 
        x = (x*x) % p;   
    } 
    return res%p;
 }
    /*public static long power(long x,long a) {
		if(a==0) return 1;
		if(a%2==1)return (x*1L*power(x,a-1))%mod;
		return (power((int)((x*1L*x)%mod),a/2))%mod;
	}*/
   static int find(int x) 
    { 
        // Finds the representative of the set 
        // that x is an element of 
        while(parent[x]!=x) 
        { 
            // if x is not the parent of itself 
            // Then x is not the representative of 
            // his set,
            x=parent[x];
  
            // so we recursively call Find on its parent 
            // and move i's node directly under the 
            // representative of this set 
        } 
  
        return x; 
    }
    static void union(int x, int y) 
    { 
        // Find representatives of two sets 
        int xRoot = find(x), yRoot = find(y); 
  
        // Elements are in the same set, no need 
        // to unite anything. 
        if (xRoot == yRoot) 
            return; 
  
         // If x's rank is less than y's rank 
        if (rank[xRoot] < rank[yRoot]) 
  
            // Then move x under y  so that depth 
            // of tree remains less 
            parent[xRoot] = yRoot; 
  
        // Else if y's rank is less than x's rank 
        else if (rank[yRoot] < rank[xRoot]) 
  
            // Then move y under x so that depth of 
            // tree remains less 
            parent[yRoot] = xRoot; 
  
        else // if ranks are the same 
        { 
            // Then move y under x (doesn't matter 
            // which one goes where) 
            parent[yRoot] = xRoot; 
  
            // And increment the the result tree's 
            // rank by 1 
            rank[xRoot] = rank[xRoot] + 1; 
        } 
    } 
    static long gcd(long a, long b) 
    { 
        if (a == 0){ 
             //ans+=b;
            return b;
        }
        return gcd(b % a, a); 
    }
}