p01970 The Diversity of Prime Factorization
D: The Diversity of Prime Factorization

Problem

Ebi-chan has the FACTORIZATION MACHINE, which can factorize natural numbers M (greater than 1) in O ($ \ log $ M) time! But unfortunately, the machine could display only digits and white spaces.

In general, we consider the factorization of M as p_1 ^ {e_1} \ times p_2 ^ {e_2} \ times ... \ times p_K ^ {e_K} where (1) i <j implies p_i <p_j and (2) p_i is prime. Now, she gives M to the machine, and the machine displays according to the following rules in ascending order with respect to i:

* If e_i = 1, then displays p_i,
* otherwise, displays p_i e_i.



For example, if she gives either `22` or` 2048`, then `2 11` is displayed. If either` 24` or `54`, then` 2 3 3`.

Okay, Ebi-chan has written down the output of the machine, but she notices that she has forgotten to write down the input! Now, your task is to count how many natural numbers result in a noted output. Note that Ebi-chan has mistaken writing and no input could result in the output.

The answer could be too large, so, you must output it modulo 10 ^ 9 + 7 (prime number).

Input


N
q_1 q_2 $ \ cdots $ q_N


In the first line, the number of the output of the machine is given. In the second line, the output of the machine is given.

Constraints

* 1 \ leq N \ leq 10 ^ 5
* 2 \ leq q_i \ leq 10 ^ 6 (1 \ leq i \ leq N)



Output

Print the number of the natural numbers that result in the given output of the machine.

Sample Input 1


3
2 3 3


Sample Output for Input 1


2

24 = 2 ^ 3 \ times 3 and 54 = 2 \ times 3 ^ 3 satisfy the condition.

Sample Input 2


3
2 3 4


Sample Output 2 for Input 2


1

Only 162 = 2 \ times 3 ^ 4 satisfies the condition. Note that 4 is not prime.

Sample Input 3


3
3 5 2


Sample Output for Input 3


1

Since 2 <3 <5, only 75 = 3 \ times 5 ^ 2 satisfies the condition.

Sample Input 4


1
Four


Sample Output for Input 4


0

Ebi-chan should have written down it more carefully.





Example

Input

3
2 3 3


Output

2
```python
from collections import defaultdict
MAX = 1000000
ROOT = 1000
MOD = 1000000007
is_prime = [True] * (MAX + 1)
is_prime[0] = is_prime[1] = False
for i in range(2, ROOT + 1):
  if is_prime[i]:
    for j in range(i * i, MAX + 1, i):
      is_prime[j] = False

n = int(input())
qlst = list(map(int, input().split()))
total1 = 0#next is kisuu or sisuu
total2 = 1#next is kisuu only(pre is index)
last_prime = 0
dic = {}
dic[(last_prime, 0)] = total1
dic[(last_prime, 1)] = total2
for q in qlst:
  new_dic = defaultdict(int)
  for k, v in dic.items():
    last_prime, t = k
    if is_prime[q]:
      if t == 0:
        if last_prime < q:
          new_dic[(q, 0)] = (new_dic[(q, 0)] + v) % MOD
          new_dic[(last_prime, 1)] = (new_dic[(last_prime, 1)] + v) % MOD
        else:
          new_dic[(last_prime, 1)] = (new_dic[(last_prime, 1)] + v) % MOD
      else:
        if last_prime < q:
          new_dic[(q, 0)] = (new_dic[(q, 0)] + v) % MOD
    
    if not is_prime[q]:
      if t == 0:
        new_dic[(last_prime, 1)] = (new_dic[(last_prime, 1)] + v) % MOD
  dic = new_dic
print(sum(dic.values()) % MOD)

```