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.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

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

    static class CBanMi {
        long mod = (long) (1e9 + 7);

        long calc(long n, long k) {

            long res = 0;

            long r1 = (Utils.pow(2, n, mod) + mod - 1) % mod;

            long a2 = r1 * (Utils.pow(2, k, mod) + mod - 1) % mod;


            return (r1 + a2) % mod;

        }

        public void solve(int testNumber, FastScanner in, PrintWriter out) {
            int n = in.readInt();
            int q = in.readInt();

            String str = in.readToken();
            SumCounterInt sc = new SumCounterInt(n);


            for (int i = 0; i < str.length(); i++) {
                sc.next(str.charAt(i) == '1' ? 1 : 0);
            }

            for (int i = 0; i < q; i++) {
                int l = in.readInt() - 1, r = in.readInt() - 1;

                long tot = (int) sc.getSum(l, r + 1);

                tot = calc(tot, r - l + 1 - tot);
                out.println(tot);
            }
        }

    }

    static class FastScanner {
        private StringTokenizer st;
        private BufferedReader in;

        public FastScanner(final InputStream in) {
            this.in = new BufferedReader(new InputStreamReader(in));
        }

        public String readToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (final IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        public int readInt() {
            return Integer.parseInt(readToken());
        }

    }

    static final class Utils {
        private Utils() {
        }

        public static long pow(long x, long n, long mod) {
            long res = 1;
            for (long p = x; n > 0; n >>= 1, p = (p * p) % mod) {
                if ((n & 1) != 0) {
                    res = res * p % mod;
                }
            }
            return res;
        }

        public static int constrain(int value, int min, int max) {
            return Math.min(Math.max(value, min), max);
        }

    }

    static class SumCounterInt {
        private int n;
        private int pos;
        private int[] sums;
        private boolean fixRanges;

        public SumCounterInt(int n) {
            this(n, true);
        }

        public SumCounterInt(int n, boolean fixRanges) {
            this.n = n;
            this.pos = 1;
            this.sums = new int[n + 1];
            this.fixRanges = fixRanges;
        }

        public void next(int value) {
            sums[pos] = sums[pos - 1] + value;
            pos++;
        }

        public long getSum(int from, int to) {
            if (fixRanges) {
                to = Utils.constrain(to, 0, n);
                from = Utils.constrain(from, 0, n - 1);

                if (from >= to) {
                    return 0;
                }
            }

            return sums[to] - sums[from];
        }

    }
}

