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:
/**
 * Created by Aminul on 11/14/2018.
 */

import java.io.*;
import java.util.*;
import static java.lang.Math.*;

public class C {
    public static void main(String[] args)throws Exception {
        FastReader in = new FastReader(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int n = in.nextInt(), q = in.nextInt();
        char [] s = in.next(n);
        int a[] = new int[n+1];
        for (int i = 1; i <= n; i++) {
            a[i] = (s[i-1] - '0') + a[i-1];
        }

        for (int i = 0; i < q; i++) {
            int l = in.nextInt(), r = in.nextInt();
            int range = (r-l+1), ones = ones(a, l, r), zero = range - ones;
            if(ones == 0) {
                pw.println(0); continue;
            }

            long N = geometricSum_2(1, ones);
            long zeroStart = mod(bigMod(2, ones, mod)-1, mod);
            long M = geometricSum_2(zeroStart, zero);

            long res = (N + M + mod) % mod;

            pw.println(res);

        }
          
        pw.close();
    }

    static long geometricSum_2(long a, long n) {
        long res = mod(bigMod(2, n, mod)-1, mod);
        res = (res * a) % mod;
        return res;
    }

    static int ones(int a[], int i, int j) {
        return a[j] - a[i-1];
    }

    static long mod = (long)1e9+7;
    static long inv2;

    static long bigMod ( long a, long p, long m ) { // returns (a^p) % m
        long res = 1 % m, x = a % m;
        while ( p > 0 ) {
            if ( (p & 1) > 0 ) res = ( res * x ) % m;
            x = ( x * x ) % m; p >>= 1;
        }
        return res;
    }

    public static long modInverse(long a, long m) {
        a = mod(a, m);
        return a == 0 ? 0 : mod((1 - modInverse(m % a, a) * m) / a, m);
    }

    public static long mod(long a, long m) {
        long A =  (a % m);
        return A >= 0 ? A : A + m;
    }

    static void debug(Object...obj) {
        System.err.println(Arrays.deepToString(obj));
    }

    static class FastReader {
        InputStream is;
        private byte[] inbuf = new byte[1024];
        private int lenbuf = 0, ptrbuf = 0;

        public FastReader(InputStream is) {
            this.is = is;
        }

        public int readByte() {
            if (lenbuf == -1) throw new InputMismatchException();
            if (ptrbuf >= lenbuf) {
                ptrbuf = 0;
                try {
                    lenbuf = is.read(inbuf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (lenbuf <= 0) return -1;
            }
            return inbuf[ptrbuf++];
        }

        public boolean isSpaceChar(int c) {
            return !(c >= 33 && c <= 126);
        }
        private boolean isEndOfLine(int c) {
            return c == '\n' || c == '\r' || c == -1;
        }

        public int skip() {
            int b;
            while ((b = readByte()) != -1 && isSpaceChar(b)) ;
            return b;
        }

        public String next() {
            int b = skip();
            StringBuilder sb = new StringBuilder();
            while (!(isSpaceChar(b))) {
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }



        public String nextLine() {
            int c = skip();
            StringBuilder sb = new StringBuilder();
            while (!isEndOfLine(c)){
                sb.appendCodePoint(c);
                c = readByte();
            }
            return sb.toString();
        }

        public int nextInt() {
            int num = 0, b;
            boolean minus = false;
            while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
            if (b == '-') {
                minus = true;
                b = readByte();
            }
            while (true) {
                if (b >= '0' && b <= '9') {
                    num = (num << 3) + (num << 1) + (b - '0');
                } else {
                    return minus ? -num : num;
                }
                b = readByte();
            }
        }

        public long nextLong() {
            long num = 0;
            int b;
            boolean minus = false;
            while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
            if (b == '-') {
                minus = true;
                b = readByte();
            }

            while (true) {
                if (b >= '0' && b <= '9') {
                    num = (num << 3) + (num << 1) + (b - '0');
                } else {
                    return minus ? -num : num;
                }
                b = readByte();
            }
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public char[] next(int n) {
            char[] buf = new char[n];
            int b = skip(), p = 0;
            while (p < n && !(isSpaceChar(b))) {
                buf[p++] = (char) b;
                b = readByte();
            }
            return n == p ? buf : Arrays.copyOf(buf, p);
        }

        public char readChar() {
            return (char) skip();
        }
    }
}