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.IOException;
import java.io.InputStreamReader;
import java.io.Reader;
import java.util.*;
public class BanhMi {
    public static final long MOD = 1000000007;
    public static void main(String[] args) {
        FastScanner scanner = new FastScanner();
        int N = scanner.nextInt();
        StringBuilder stringBuilder = new StringBuilder();
        int Q = scanner.nextInt();
        BIT bit = new BIT(N+1);
        char[] vals = scanner.next().toCharArray();
        for(int i = 1; i <= N; i++) {
            bit.update(i, vals[i-1]-'0');
        }
        for(int i = 0; i < Q; i++) {
            int l = scanner.nextInt();
            int r = scanner.nextInt();
            stringBuilder.append(calcVal((int)bit.rangeSum(l, r), r-l+1)).append("\n");
        }
        System.out.print(stringBuilder.toString());
    }
    public static long calcVal(int val, int sz) {
        if (val == 0) return 0;
        int nZeroes = sz - val;
        long ret = (modPow(2, val-1)*2-1)%MOD;
        ret = (ret*modPow(2, nZeroes))%MOD;
        return ret;
    }
    public static long modPow(long x, long y) {
        if (y == 0) return 1;
        long p = modPow(x, y / 2) %MOD;
        p = (p * p) % MOD;
        if (y % 2 == 0) return p;
        else return (x * p) % MOD;
    }
    public static class BIT {
        long[] table;
        int sz;
        public BIT(int ss){
            table = new long[ss];
            sz = ss;
        }
        public void update(int loc, long amt) {
            while(loc < sz) {
                table[loc]+=amt;
                loc += Integer.lowestOneBit(loc);
            }
        }
        public long prefixSum(int loc) {
            long ret = 0L;
            while(loc > 0) {
                ret+=table[loc];
                loc-=Integer.lowestOneBit(loc);
            }
            return ret;
        }
        public long rangeSum(int l, int r) {
            return prefixSum(r) - prefixSum(l-1);
        }
    }
    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        public FastScanner(Reader in) {
            br = new BufferedReader(in);
        }
        public FastScanner() {
            this(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 readNextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
        int[] readIntArray(int n) {
            int[] a = new int[n];
            for (int idx = 0; idx < n; idx++) {
                a[idx] = nextInt();
            }
            return a;
        }
    }
}