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.
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.DataInputStream;
import java.io.IOException;
import java.util.InputMismatchException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author xwchen
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int q = in.nextInt();
            char[] s = in.next().toCharArray();
            int[] sum = new int[s.length + 1];
            long[] e2 = new long[s.length + 1];
            long mod = 1000000007L;
            e2[0] = 1;
            for (int i = 1; i <= n; ++i) {
                sum[i] = sum[i - 1] + ((s[i - 1] == '1') ? 1 : 0);
                e2[i] = e2[i - 1] * 2;
                e2[i] %= mod;
            }


            while (q-- > 0) {
                int l = in.nextInt();
                int r = in.nextInt();

                int len = r - l + 1;
                int one = sum[r] - sum[l - 1];
                int zero = len - one;
                long ans = e2[one] - 1;
                long base = ans;
                ans = ans + (e2[zero] - 1) * base % mod;
                ans %= mod;
                out.println(ans);
            }
        }

    }

    static class InputReader {
        final private int BUFFER_SIZE = 1 << 10;
        final private int LINE_SIZE = 1 << 20;
        private DataInputStream in;
        private byte[] buffer;
        private int bufferPointer;
        private int bytesRead;

        public InputReader(InputStream inputStream) {
            this.in = new DataInputStream(inputStream);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String next() {
            byte[] buf = new byte[LINE_SIZE]; // line length
            int cnt = 0, c;
            c = read();
            while (c == ' ' || c == '\n' || c == '\r')
                c = read();
            do {
                if (c == ' ' || c == '\n' || c == '\r')
                    break;
                buf[cnt++] = (byte) c;
            } while ((c = read()) != -1);
            return new String(buf, 0, cnt);
        }

        public int nextInt() {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (neg)
                return -ret;
            return ret;
        }

        private void fillBuffer() {
            try {
                bytesRead = in.read(buffer, bufferPointer = 0, BUFFER_SIZE);
                if (bytesRead == -1)
                    buffer[0] = -1;
            } catch (IOException e) {
                throw new InputMismatchException();
            }
        }

        private byte read() {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }

    }
}

