Codeforces Round #445 C. Petya and Catacombs【思维/题意】

C. Petya and Catacombs

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A very brave explorer Petya once decided to explore Paris catacombs. Since Petya is not really experienced, his exploration is just walking through the catacombs.

Catacombs consist of several rooms and bidirectional passages between some pairs of them. Some passages can connect a room to itself and since the passages are built on different depths they do not intersect each other. Every minute Petya arbitrary chooses a passage from the room he is currently in and then reaches the room on the other end of the passage in exactly one minute. When he enters a room at minute i, he makes a note in his logbook with number ti:

• If Petya has visited this room before, he writes down the minute he was in this room last time;
• Otherwise, Petya writes down an arbitrary non-negative integer strictly less than current minute i.

Initially, Petya was in one of the rooms at minute 0, he didn't write down number t0.

At some point during his wandering Petya got tired, threw out his logbook and went home. Vasya found his logbook and now he is curious: what is the minimum possible number of rooms in Paris catacombs according to Petya's logbook?

Input

The first line contains a single integer n (1 ≤ n ≤ 2·105) — then number of notes in Petya's logbook.

The second line contains n non-negative integers t1, t2, ..., tn (0 ≤ ti < i) — notes in the logbook.

Output

In the only line print a single integer — the minimum possible number of rooms in Paris catacombs.

Examples

input

2
0 0

output

2

input

5
0 1 0 1 3

output

3

Note

In the first sample, sequence of rooms Petya visited could be, for example 1 → 1 → 2, 1 → 2 → 1 or 1 → 2 → 3. The minimum possible number of rooms is 2.

In the second sample, the sequence could be 1 → 2 → 3 → 1 → 2 → 1.

【题意】：相同房间写下上一次到达的时间点，不同房间是随机一下，因为题目要最小答案 。

【分析】：每个时间点只会出现一次，出现多次说明是随机的，也就是说出现多次的那些说明是不同房间，那么只要标记出现过的 然后统计再次出现的次数。0这个点也算。

【代码】：

#include <bits/stdc++.h>

using namespace std;

const int maxn = 1e5*2;
int a[maxn];


int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
    }
    sort(a,a+n);
    int cnt = 1;
    for(int i=1;i<n;i++)
    {
        if(a[i]==a[i-1])
            cnt++;
    }
    cout<<cnt<<endl;
    return 0;
}