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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <inttypes.h>
#include <string.h>
#include <math.h>

int main (int argc, char const *argv[])
{
    intmax_t N, M, a[(int)1e5];
    intmax_t c[(int)1e5], ans = 1;

    scanf("%jd%jd", &N, &M);
    for (size_t i = 0; i < M; i++) {
        scanf("%jd", &a[i]);
    }

    c[0] = 0;
    c[1] = 1;
    for (size_t i = 2; i < N; i++) {
        c[i] = (c[i - 2] + c[i - 1]) % ((int)1e9 + 7);
    }

    if (M == 0) {
        printf("%jd\n", c[N]);
        return 0;
    }
    
    for (size_t i = 0; i <= M; i++) {
        if (i == 0) {
            ans *= c[a[i]];
        } else if (i == M) {
            ans *= c[N - a[i - 1]];
        } else {
            ans *= c[a[i] - a[i - 1] - 1];
        }
        ans %= (int)1e9 + 7;
    }

    printf("%jd\n", ans);

    return 0;
}