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.text.*;
import java.util.*;
import java.io.*;
public class Main {
    static final int mod = 1000000007;
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer stk = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(stk.nextToken());
        int q = Integer.parseInt(stk.nextToken());
        String input = br.readLine();
        ArrayList<Integer> d = new ArrayList<>();
        for(int i=0;i<n;i++){
            d.add(input.charAt(i)-'0');
        }
        SegmentTree st = new SegmentTree(n);
        st.insert(d);
        StringBuilder sb = new StringBuilder();
        for(int i=0;i<q;i++){
            stk = new StringTokenizer(br.readLine());
            int l = Integer.parseInt(stk.nextToken());
            int r = Integer.parseInt(stk.nextToken());
            int sum = st.getSum(l,r);
            sb.append(getBinaryToDecimal(r-l+1,sum)).append('\n');
        }
        System.out.print(sb);
    }
    public static long getBinaryToDecimal(int len,int one){
        int zero = len-one;
        return (getPow(2,len)-getPow(2,zero)+mod)%mod;
    }
    public static long getPow(long n,long exp){
        if(exp == 0 ) return 1;
        if(exp == 1) return n%mod;
        if(exp %2 ==0){
            long v = getPow(n,exp/2);
            return v*v%mod;
        }
        else{
            long v = getPow(n,exp-1);
            return v*n%mod;
        }
    }
}

class SegmentTree {
    ArrayList<Integer> tree;
    int s;

    public SegmentTree(int n) {
        for (s = 1; s < n; s *= 2);
        tree = new ArrayList<Integer>(s * 2);
        tree.add(0);
        for (int i = 1; i < s + s; i++)
            tree.add(0);
    }

    void insert(ArrayList<Integer> d) {
        for (int i = s; i < s + d.size(); i++)
            tree.set(i, d.get(i - s));
        for (int i = s - 1; i >= 1; i--)
            tree.set(i, (tree.get(i * 2) + tree.get(i * 2 + 1)));
    }

    int getSum(int Left, int Right) {
        int l = Left + s - 1, r = Right + s - 1;
        int rval = 0;
        while (l <= r) {
            if (l % 2 == 0) l /= 2;
            else {
                rval += tree.get(l);
                l = (l / 2) + 1;
            }
            if (r % 2 == 1) r /= 2;
            else {
                rval += tree.get(r);
                r = (r / 2) - 1;
            }
        }
        return rval;
    }
}