Problem:
Score : 300 points

Problem Statement

There is a staircase with N steps. Takahashi is now standing at the foot of the stairs, that is, on the 0-th step.
He can climb up one or two steps at a time.

However, the treads of the a_1-th, a_2-th, a_3-th, \ldots, a_M-th steps are broken, so it is dangerous to set foot on those steps.

How many are there to climb up to the top step, that is, the N-th step, without setting foot on the broken steps?
Find the count modulo 1\ 000\ 000\ 007.

Constraints

1 \leq N \leq 10^5

0 \leq M \leq N-1

1 \leq a_1 < a_2 < ... < a_M \leq N-1

Input

Input is given from Standard Input in the following format:

N M
a_1
a_2
.
.
.
a_M

Output

Print the number of ways to climb up the stairs under the condition, modulo 1\ 000\ 000\ 007.

Sample Input 1

6 1
3

Sample Output 1

4

There are four ways to climb up the stairs, as follows:

0 \to 1 \to 2 \to 4 \to 5 \to 6

0 \to 1 \to 2 \to 4 \to 6

0 \to 2 \to 4 \to 5 \to 6

0 \to 2 \to 4 \to 6

Sample Input 2

10 2
4
5

Sample Output 2

0

There may be no way to climb up the stairs without setting foot on the broken steps.

Sample Input 3

100 5
1
23
45
67
89

Sample Output 3

608200469

Be sure to print the count modulo 1\ 000\ 000\ 007.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.*;

public class Main {

    public static void main(String[] args) {
//        long startTime = System.currentTimeMillis();
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        int M = scanner.nextInt();
        List<Integer> aList = new ArrayList<>();
        for (int i = 0; i < M; i++) {
            aList.add(scanner.nextInt());
        }
        scanner.close();
        long count = 0;
        long calcValue = 1000000007;

        Map<Integer, Long> map = new HashMap<>();
        map.put(0, 1L);
//        Queue<Integer> queue = new ArrayDeque<>();
//        queue.add(0);
//        System.out.println("while" + queue);
        for (int i = 0; i < N; i++) {
            int pollValue = i;
            if (aList.contains(i)) {
                continue;
            }

            long valueCount = map.get(pollValue);
            // 1段上がる場合
            int one = pollValue + 1;
            if (one <= N) {
                if (one == N) {
                    count += valueCount;
                } else if (!aList.contains(one)) {
                    if (map.containsKey(one)) {
                        long oneCount = map.get(one);
                        oneCount += (valueCount);
                        map.put(one, oneCount);
                    } else {
                        map.put(one, valueCount);
                    }
                }
            }
            // 2段上がる場合
            int two = pollValue + 2;
            if (two <= N) {
                if (two == N) {
                    count += valueCount;
                } else if (!aList.contains(two)) {
                    if (map.containsKey(two)) {
                        long twoCount = map.get(two);
                        twoCount += (valueCount);
                        map.put(two, twoCount);
                    } else {
                        map.put(two, valueCount);
                    }
                }
            }
            count = count % calcValue;
        }

//        while (!queue.isEmpty()) {
//            int pollValue = queue.poll();
//            long valueCount = map.get(pollValue);
//            // 1段上がる場合
//            int one = pollValue + 1;
//            if (one <= N) {
//                if (one == N) {
//                    count += valueCount;
//                } else if (!aList.contains(one)) {
//                    if (map.containsKey(one)) {
//                        long oneCount = map.get(one);
//                        oneCount += (valueCount);
//                        map.put(one, oneCount);
//                    } else {
//                        queue.add(one);
//                        map.put(one, valueCount);
//                    }
//                }
//            }
//            // 2段上がる場合
//            int two = pollValue + 2;
//            if (two <= N) {
//                if (two == N) {
//                    count += valueCount;
//                } else if (!aList.contains(two)) {
//                    if (map.containsKey(two)) {
//                        long twoCount = map.get(two);
//                        twoCount += (valueCount);
//                        map.put(two, twoCount);
//                    } else {
//                        queue.add(two);
//                        map.put(two, valueCount);
//                    }
//                }
//            }
//            count = count % calcValue;
////            System.out.println("while" + queue);
//        }

//        map.entrySet().forEach(entry -> System.out.println(entry.getKey() + " / " + entry.getValue()));

        System.out.println(count);

//        System.out.println(System.currentTimeMillis() - startTime);
    }

}