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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

// ぬけのないときをフィボッチ数列に則る
// 前置代入
int fib(int n) {
  if (n == 0) return 1;
  if (n == 1) return 1;
  return fib(n-1) + fib(n-2);
}

int main() {
  int n, m;
  int sol = 1;
  scanf("%d", &n);
  scanf("%d", &m);
  int am, prevam = 0;
  // 連続した階段が壊れているとき0を出力して終了
  int i = 0;
  while (i++ < m) {
    scanf("%d", &am);
    if (prevam != 0 && prevam + 1 == am) {
      printf("0\n");
      return 0;
    }
    sol *= fib(am - prevam - 1);
    prevam = am;
    am = 0;
  }
  sol *= fib(n-prevam-1);
  printf("%d\n", sol%1000000007);
  return 0;
}