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: Accepted)

#include <stdio.h>
#define MEM 100000

int main(void) {
    int N, M, broken[MEM], b = 0;
    long stairs[MEM];

    scanf("%d %d", &N, &M);
    for (int i = 0; i < M; i++) scanf("%d", broken+i);

    if (0 != (broken[b]-1)) {
        stairs[0] = 1;
    } else {
        stairs[0] = 0;
        b++;
    }
    if (1 != (broken[b]-1)) {
        stairs[1] = 1 + stairs[0];
    } else {
        stairs[1] = 0;
        b++;
    }

    for (int i = 2; i < N; i++) {
        if (i != broken[b]-1) {
            stairs[i] = stairs[i-1] + stairs[i-2];
            if(stairs[i] >= 1000000007) stairs[i] %= 1000000007;
        } else {
            stairs[i] = 0;
            b++;
        }
    }

    printf("%d\n", stairs[N-1]);
    return 0;
}