1096_F. Inversion Expectation
A permutation of size n is an array of size n such that each integer from 1 to n occurs exactly once in this array. An inversion in a permutation p is a pair of indices (i, j) such that i > j and a_i < a_j. For example, a permutation [4, 1, 3, 2] contains 4 inversions: (2, 1), (3, 1), (4, 1), (4, 3).

You are given a permutation p of size n. However, the numbers on some positions are replaced by -1. Let the valid permutation be such a replacement of -1 in this sequence back to numbers from 1 to n in such a way that the resulting sequence is a permutation of size n.

The given sequence was turned into a valid permutation randomly with the equal probability of getting each valid permutation.

Calculate the expected total number of inversions in the resulting valid permutation.

It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q ≠ 0. Report the value of P ⋅ Q^{-1} \pmod {998244353}.

Input

The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the sequence.

The second line contains n integers p_1, p_2, ..., p_n (-1 ≤ p_i ≤ n, p_i ≠ 0) — the initial sequence.

It is guaranteed that all elements not equal to -1 are pairwise distinct.

Output

Print a single integer — the expected total number of inversions in the resulting valid permutation.

It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q ≠ 0. Report the value of P ⋅ Q^{-1} \pmod {998244353}.

Examples

Input


3
3 -1 -1


Output


499122179


Input


2
1 2


Output


0


Input


2
-1 -1


Output


499122177

Note

In the first example two resulting valid permutations are possible:

  * [3, 1, 2] — 2 inversions; 
  * [3, 2, 1] — 3 inversions. 



The expected value is (2 ⋅ 1 + 3 ⋅ 1)/(2) = 2.5.

In the second example no -1 are present, thus the only valid permutation is possible — the given one. It has 0 inversions.

In the third example there are two resulting valid permutations — one with 0 inversions and one with 1 inversion.
```python
base=998244353;
def power(x, y):
    if(y==0):
        return 1
    t=power(x, y//2)
    t=(t*t)%base
    if(y%2):
        t=(t*x)%base
    return t;
def inverse(x):
    return power(x, base-2)
ft=[0]
for i in range(0, 200000):
    ft.append(0)
def get(i):
    res=0
    while(i<=200000):
        res+=ft[i]
        i+=i&-i
    return res
def update(i, x):
    while(i):
        ft[i]+=x
        i-=i&-i
n=int(input())
a=[0]
a+=list(map(int, input().split()))
neg=[0]
non=[0]
for i in range(1, n+1):
    non.append(0)
for i in range(1, n+1):
    if(a[i]!=-1):
        non[a[i]]+=1
for i in range(1, n+1):
    non[i]+=non[i-1]
for i in range(1, n+1):
    if(a[i]==-1):
        neg.append(neg[i-1]+1)
    else: 
        neg.append(neg[i-1])
m=neg[n]
ans=0
for i in range(1, n+1):
    if(a[i]!=-1):
        ans+=get(a[i])
        update(a[i], 1)
fm=1
fs=fm
for i in range(1, m+1):
    fs=fm
    fm=(fm*i)%base
fs=(fs*inverse(fm))%base
for i in range(1, n+1):
    if(a[i]!=-1):
        less=a[i]-non[a[i]]
        more=m-less
        ans=(ans+neg[i]*more*fs)%base
        ans=(ans+(m-neg[i])*less*fs)%base
ans=(ans+m*(m-1)*inverse(4))%base
print(ans)
```