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: Accepted)

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

#define N_MAX (100)
#define P_MAX (100)
#define DP_ARRAY_SIZE (N_MAX * P_MAX / 32 + 1)

#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define ABS(a) ((a) < 0 ? -(a) : (a))
#define ABSS(a, b) ((a) > (b) ? (a) - (b) : (b) - (a))

int compare_sz_desc(const void* a, const void* b) {
    return *((size_t*)a) > * ((size_t*)b) ? -1 : 1;
}

int compare_f(const void* a, const void* b) {
    return *((int64_t*)a) < *((int64_t*)b) ? -1 : 1;
}

int compare_f2(const void* a, const void* b) {
    return *((uint64_t*)a) > * ((uint64_t*)b) ? -1 : 1;
}

static size_t comb(const size_t n, const size_t r) {
    size_t result = 1;

    for (size_t i = 0; i < r; i++) {
        result *= n - i;
        result /= i + 1;
    }

    return result;
}

static uint64_t combU64(const uint64_t n, const uint64_t r) {
    uint64_t result = 1;

    for (uint64_t i = 0; i < r; i++) {
        result *= n - i;
        result /= i + 1;
    }

    return result;
}

static uint64_t gcdU64(uint64_t m, uint64_t n)
{
    uint64_t temp;

    while (m % n != 0) {
        temp = n;
        n = m % n;
        m = temp;
    }

    return n;
}

static size_t gcdZu(size_t m, size_t n) {
    size_t temp;

    while (m % n != 0) {
        temp = n;
        n = m % n;
        m = temp;
    }

    return n;
}

static size_t a[100000];

int main(void) {
    size_t N, M;

    scanf("%zu %zu\n", &N, &M);
    for (size_t i = 0; i < M; i++) {
        scanf("%zu", &(a[i]));
        if (i < M - 1) {
            scanf("\n");
        }
    }

    size_t dp[2];
    dp[0] = 1;
    size_t pos = 0;
    for (size_t i = 1; i <= N; i++) {
        if (i == 1) {
            if (a[pos] == i) {
                dp[1] = 0;
                pos++;
                while (true) {
                    if (pos < M - 1 && a[pos] == a[pos - 1]) {
                        pos++;
                    }
                    else {
                        break;
                    }
                }
            }
            else {
                dp[1] = 1;
            }
        }
        else {
            if (a[pos] == i) {
                dp[i%2] = 0;
                pos++;
                while (true) {
                    if (pos < M - 1 && a[pos] == a[pos - 1]) {
                        pos++;
                    }
                    else {
                        break;
                    }
                }
            }
            else {
                dp[i % 2] = (dp[0] + dp[1]) % 1000000007;
            }
        }
    }

    printf("%zu\n", dp[N%2]);

    return 0;
}