Codeforces Round #445 C. Petya and Catacombs（思维）

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：有位置个房间两两相通。0时刻，一个在人任意一个房间里，每一分种，这个人会选则走进其他房间或者留在这个房间。如果这个人是第一次来到这个房间，他会随便写下一个小于当前时间的数，如果他是不是第一次来到这个房间，他就会写下他上一次来到这个房间的时间。（注意，如果开始在房间1,1分钟后，他选择留在房间1，那么视他是第二次进入该房间）。现在，给出他记录的n个数字，求最少有多少个房间。

 

思路：不管怎么记录，如果同样数字记录了多次，那么这些数字必定是在不同房间里面记录的。也就是说，我们只要记下重复的数字的个数就好了。刚开始在0时刻，这个人也处于一个房间，所以最后加1。

 

#include "iostream"
#include "algorithm"
using namespace std;
const int Max = 2e5+10;
int a[Max];
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    sort(a,a+n);
    int cnt=0;
    for(int i=1;i<n;i++){
        if(a[i]==a[i-1])
            cnt++;
    }
    cout<<cnt+1<<endl;
    return 0;
}