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>
int main(void){
    // Your code here!
    int i, j, n, m;
    int dp[100002];
    int flag;
    int a[100002];
    int js = 0;

    scanf("%d %d", &n, &m);

    for (i = 0; i < m; i++) {
        scanf("%d", &a[i]);
    }
    dp[0] = 1;

    flag = 0;
    for (j = 0; j < m; j++) {
        if (1 == a[j]) {
            flag = 1;
            break;
        }
    }
    if (flag) {
        dp[1] = 0;
    } else {
        dp[1] = 1;
    }


    for (i = 2; i <= n; i++) {
        flag = 0;
        for (j = js; j < m; j++) {
            if (i == a[j]) {
                flag = 1;
                js = j + 1;
                break;
            }
        }
        if (flag) {
            dp[i] = 0;
            continue;
        }
        dp[i] = (dp[i - 1] + dp[i - 2]) % 1000000007;
    }
    printf("%d\n", dp[n]);

    return 0;
}