341_C. Iahub and Permutations
Iahub is so happy about inventing bubble sort graphs that he's staying all day long at the office and writing permutations. Iahubina is angry that she is no more important for Iahub. When Iahub goes away, Iahubina comes to his office and sabotage his research work.

The girl finds an important permutation for the research. The permutation contains n distinct integers a1, a2, ..., an (1 ≤ ai ≤ n). She replaces some of permutation elements with -1 value as a revenge. 

When Iahub finds out his important permutation is broken, he tries to recover it. The only thing he remembers about the permutation is it didn't have any fixed point. A fixed point for a permutation is an element ak which has value equal to k (ak = k). Your job is to proof to Iahub that trying to recover it is not a good idea. Output the number of permutations which could be originally Iahub's important permutation, modulo 1000000007 (109 + 7).

Input

The first line contains integer n (2 ≤ n ≤ 2000). On the second line, there are n integers, representing Iahub's important permutation after Iahubina replaces some values with -1. 

It's guaranteed that there are no fixed points in the given permutation. Also, the given sequence contains at least two numbers -1 and each positive number occurs in the sequence at most once. It's guaranteed that there is at least one suitable permutation.

Output

Output a single integer, the number of ways Iahub could recover his permutation, modulo 1000000007 (109 + 7).

Examples

Input

5
-1 -1 4 3 -1


Output

2

Note

For the first test example there are two permutations with no fixed points are [2, 5, 4, 3, 1] and [5, 1, 4, 3, 2]. Any other permutation would have at least one fixed point. 
```python
#lahub and Permutations
import sys
readline = sys.stdin.buffer.readline
def even(n): return 1 if n%2==0 else 0
mod = 10**9+7
def pow(n,p,mod=mod): #繰り返し二乗法(nのp乗)
    res = 1
    while p > 0:
        if p % 2 == 0:
            n = n ** 2 % mod
            p //= 2
        else:
            res = res * n % mod
            p -= 1
    return res % mod

def factrial_memo(n=10**5,mod=mod):
    fact = [1, 1]
    for i in range(2, n + 1):
        fact.append((fact[-1] * i) % mod)
    return fact

fact = factrial_memo()

def permutation(n,r): #nPr
    return fact[n]*pow(fact[n-r],mod-2)%mod
def combination(n,r): #nCr
    return permutation(n,r)*pow(fact[r],mod-2)%mod
    #return fact[n]*pow(fact[n-r],mod-2)*pow(fact[r],mod-2)
def homogeneous(n,r): #nHr
    return combination(n+r-1,r)%mod
    #return fact[n+m-1]*pow(fact[n-1],mod-2)*pow(fact[r],mod-2)

n = int(readline())
lst1 = list(map(int,readline().split()))
ct = 0
lst2 = [0]*2_001
lst3 = [0]*2_001
for i in range(n):
    if i+1 == lst1[i]:
        print(0)
        exit()
    if lst1[i] == -1:
        ct += 1
        lst3[i+1] = 1
    else:
        lst2[lst1[i]] = 1

ct2 = 0
for i in range(1,n+1):
    if lst3[i] == 1 and lst2[i] == 0:
        ct2 += 1
#lst2:その場所が埋まってないindex
#lst3:その数字が使われてるindex
#何個入れちゃいけない位置に入れるかで数え上げる
#入れちゃいけないものはct2個って、
#ct-ct2個のものはどこに入れても要件を満たす

ans = 0
for i in range(ct2+1):
    ans += pow(-1,i)*combination(ct2,i)*fact[ct-i]
    ans %= mod
print(ans)
```