CodeForces - 931D Peculiar apple-tree

D. Peculiar apple-tree

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

In Arcady's garden there grows a peculiar apple-tree that fruits one time per year. Its peculiarity can be explained in following way: there are n inflorescences, numbered from 1 to n. Inflorescence number 1 is situated near base of tree and any other inflorescence with number i (i > 1) is situated at the top of branch, which bottom is pi-th inflorescence and pi < i.

Once tree starts fruiting, there appears exactly one apple in each inflorescence. The same moment as apples appear, they start to roll down along branches to the very base of tree. Each second all apples, except ones in first inflorescence simultaneously roll down one branch closer to tree base, e.g. apple in a-th inflorescence gets to pa-th inflorescence. Apples that end up in first inflorescence are gathered by Arcady in exactly the same moment. Second peculiarity of this tree is that once two apples are in same inflorescence they annihilate. This happens with each pair of apples, e.g. if there are 5 apples in same inflorescence in same time, only one will not be annihilated and if there are 8 apples, all apples will be annihilated. Thus, there can be no more than one apple in each inflorescence in each moment of time.

Help Arcady with counting number of apples he will be able to collect from first inflorescence during one harvest.

Input

First line of input contains single integer number n (2 ≤ n ≤ 100 000)  — number of inflorescences.

Second line of input contains sequence of n - 1 integer numbers p2, p3, ..., pn (1 ≤ pi < i), where pi is number of inflorescence into which the apple from i-th inflorescence rolls down.

Output

Single line of output should contain one integer number: amount of apples that Arcady will be able to collect from first inflorescence during one harvest.

Examples

input

Copy

3
1 1

output

1

input

Copy

5
1 2 2 2

output

3

input

Copy

18
1 1 1 4 4 3 2 2 2 10 8 9 9 9 10 10 4

output

4

Note

In first example Arcady will be able to collect only one apple, initially situated in 1st inflorescence. In next second apples from 2nd and 3rd inflorescences will roll down and annihilate, and Arcady won't be able to collect them.

In the second example Arcady will be able to collect 3 apples. First one is one initially situated in first inflorescence. In a second apple from 2nd inflorescence will roll down to 1st (Arcady will collect it) and apples from 3rd, 4th, 5th inflorescences will roll down to 2nd. Two of them will annihilate and one not annihilated will roll down from 2-nd inflorescence to 1st one in the next second and Arcady will collect it.

题意：

一棵苹果树，有n个花蕊。花蕊成树状分布。一天果子全部成熟了，于是开始下落，每秒下落一个节点。神奇之处在于同一个节点同一时刻的果子会两两相消，即只能存k%2个果子。求可以落到地面多少果子。输入每个花蕊的前驱。

思路：

因为落的时候是一层一层的落，并且根节点只有一个1，我们就可以直接dfs求出每一层有多少果子。输出每层果子数%2之和即可。

代码：

#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
using namespace std;
#define inf 1<<29
vector<int> G[100005];
int n,dis[100005];
void dfs(int u,int step)
{
    for(int i=0;i<G[u].size();i++)
    {
        dis[step]++;
        dfs(G[u][i],step+1);
    }
}
int main()
{
    scanf("%d",&n);
    int d;
    for(int i=1;i<=n;i++)
        G[i].clear();
    memset(dis,0,sizeof dis);
    for(int i=2;i<=n;i++)
    {
        scanf("%d",&d);
        G[d].push_back(i);
    }
    dfs(1,1);
    int ans=1;
    for(int i=1;i<=n;i++)
        ans+=dis[i]%2;
    printf("%d\n",ans);
    return 0;
}