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.*;
import java.math.*;
import java.util.*;
import java.util.stream.*;
import static java.lang.Math.abs;
import static java.lang.Math.min;
import static java.lang.Math.max;
import static java.lang.Math.sqrt;

@SuppressWarnings("unchecked")
public class P1062C {

  final static BigInteger TWO = BigInteger.valueOf(2);
  final static BigInteger MOD = BigInteger.valueOf(1_000_000_007);

  BigInteger modSum(BigInteger b, BigInteger e) {
    return b.multiply(TWO.modPow(e, MOD)).subtract(b);
  }

  public void run() throws Exception {
    int n = nextInt() + 1, q = nextInt(), s [] = new int [n];
    String str = next();
    for (int i = 1; i < n; s[i] = s[i - 1] + (str.charAt(i - 1) - '0'), i++);

    while ((q--) > 0) {
      int l = nextInt() - 1, r = nextInt(), oc = s[r] - s[l], zc = r - l - oc;

      BigInteger u1 = modSum(BigInteger.ONE, BigInteger.valueOf(oc));
      BigInteger u0 = modSum(u1, BigInteger.valueOf(zc));
      println(u1.add(u0).mod(MOD));
    }
  }

  public static void main(String... args) throws Exception {
    br = new BufferedReader(new InputStreamReader(System.in));
    pw = new PrintWriter(new BufferedOutputStream(System.out));
    new P1062C().run();
    br.close();
    pw.close();
    System.err.println("\n[Time : " + (System.currentTimeMillis() - startTime) + " ms]");
  }

  static long startTime = System.currentTimeMillis();
  static BufferedReader br;
  static PrintWriter pw;
  StringTokenizer stok;

  String nextToken() throws IOException {
    while (stok == null || !stok.hasMoreTokens()) {
      String s = br.readLine();
      if (s == null) { return null; }
      stok = new StringTokenizer(s);
    }
    return stok.nextToken();
  }

  void print(byte b) { print("" + b); }
  void print(int i) { print("" + i); }
  void print(long l) { print("" + l); }
  void print(double d) { print("" + d); }
  void print(char c) { print("" + c); }
  void print(Object o) {
    if (o instanceof int[]) { print(Arrays.toString((int [])o));
    } else if (o instanceof long[]) { print(Arrays.toString((long [])o));
    } else if (o instanceof char[]) { print(Arrays.toString((char [])o));
    } else if (o instanceof byte[]) { print(Arrays.toString((byte [])o));
    } else if (o instanceof short[]) { print(Arrays.toString((short [])o));
    } else if (o instanceof boolean[]) { print(Arrays.toString((boolean [])o));
    } else if (o instanceof float[]) { print(Arrays.toString((float [])o));
    } else if (o instanceof double[]) { print(Arrays.toString((double [])o));
    } else if (o instanceof Object[]) { print(Arrays.toString((Object [])o));
    } else { print("" + o); }
  }
  void print(String s) { pw.print(s); }
  void println() { println(""); }
  void println(byte b) { println("" + b); }
  void println(int i) { println("" + i); }
  void println(long l) { println("" + l); }
  void println(double d) { println("" + d); }
  void println(char c) { println("" + c); }
  void println(Object o) { print(o); println(); }
  void println(String s) { pw.println(s); }
  int nextInt() throws IOException { return Integer.parseInt(nextToken()); }
  long nextLong() throws IOException { return Long.parseLong(nextToken()); }
  double nextDouble() throws IOException { return Double.parseDouble(nextToken()); }
  char nextChar() throws IOException { return (char) (br.read()); }
  String next() throws IOException { return nextToken(); }
  String nextLine() throws IOException { return br.readLine(); }
  int [] readInt(int size) throws IOException {
    int [] array = new int [size];
    for (int i = 0; i < size; i++) { array[i] = nextInt(); }
    return array;
  }
  long [] readLong(int size) throws IOException {
    long [] array = new long [size];
    for (int i = 0; i < size; i++) { array[i] = nextLong(); }
    return array;
  }
  double [] readDouble(int size) throws IOException {
    double [] array = new double [size];
    for (int i = 0; i < size; i++) { array[i] = nextDouble(); }
    return array;
  }
  String [] readLines(int size) throws IOException {
    String [] array = new String [size];
    for (int i = 0; i < size; i++) { array[i] = nextLine(); }
    return array;
  }

  int gcd(int a, int b) {
    if (a == 0) return Math.abs(b); if (b == 0) return Math.abs(a);
    a = Math.abs(a); b = Math.abs(b);
    int az = Integer.numberOfTrailingZeros(a), bz = Integer.numberOfTrailingZeros(b);
    a >>>= az; b >>>= bz;
    while (a != b) {
      if (a > b) { a -= b; a >>>= Integer.numberOfTrailingZeros(a); }
            else { b -= a; b >>>= Integer.numberOfTrailingZeros(b); }
    }
    return (a << Math.min(az, bz));
  }

  long gcd(long a, long b) {
    if (a == 0) return Math.abs(b); if (b == 0) return Math.abs(a);
    a = Math.abs(a); b = Math.abs(b);
    int az = Long.numberOfTrailingZeros(a), bz = Long.numberOfTrailingZeros(b);
    a >>>= az; b >>>= bz;
    while (a != b) {
      if (a > b) { a -= b; a >>>= Long.numberOfTrailingZeros(a); }
            else { b -= a; b >>>= Long.numberOfTrailingZeros(b); }
    }
    return (a << Math.min(az, bz));
  }

  void shuffle(int [] a) {
    Random r = new Random();
    for (int i = a.length - 1; i >= 0; i--) {
      int j = r.nextInt(a.length);
      int t = a[i]; a[i] = a[j]; a[j] = t;
    }
  }

  void shuffle(long [] a) {
    Random r = new Random();
    for (int i = a.length - 1; i >= 0; i--) {
      int j = r.nextInt(a.length);
      long t = a[i]; a[i] = a[j]; a[j] = t;
    }
  }

  void shuffle(Object [] a) {
    Random r = new Random();
    for (int i = a.length - 1; i >= 0; i--) {
      int j = r.nextInt(a.length);
      Object t = a[i]; a[i] = a[j]; a[j] = t;
    }
  }

  void flush() {
    pw.flush();
  }

  void pause() {
    flush();
    System.console().readLine();
  }
}