Codeforces Round #275 (Div. 1)D（树形DP）

D. Random Function and Tree

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You have a rooted tree consisting of n vertices. Let's number them with integers from 1 to n inclusive. The root of the tree is the vertex 1. For each i > 1 direct parent of the vertex i is pi. We say that vertex i is child for its direct parent pi.

You have initially painted all the vertices with red color. You like to repaint some vertices of the tree. To perform painting you use the function paint that you call with the root of the tree as an argument. Here is the pseudocode of this function:

count = 0 // global integer variable 

rnd() { // this function is used in paint code
    return 0 or 1 equiprobably
}

paint(s) {
    if (count is even) then paint s with white color
    else paint s with black color

    count = count + 1
    
    if rnd() = 1 then children = [array of vertex s children in ascending order of their numbers]
    else children = [array of vertex s children in descending order of their numbers]

    for child in children { // iterating over children array
        if rnd() = 1 then paint(child) // calling paint recursively
    }
}

As a result of this function, some vertices may change their colors to white or black and some of them may remain red.

Your task is to determine the number of distinct possible colorings of the vertices of the tree. We will assume that the coloring is possible if there is a nonzero probability to get this coloring with a single call of paint(1). We assume that the colorings are different if there is a pair of vertices that are painted with different colors in these colorings. Since the required number may be very large, find its remainder of division by 1000000007 (109 + 7).

Input

The first line contains a single integer n (2 ≤ n ≤ 105) — the number of vertexes in the tree.

The second line contains n - 1 integers p2, p3, ..., pn (1 ≤ pi < i). Number pi is the parent of vertex i.

Output

Print a single integer — the answer to the problem modulo 1000000007 (109 + 7)

Sample test(s)

input

4
1 2 1

output

8

input

3
1 1

output

5

Note

All possible coloring patterns of the first sample are given below.

题意：RT

思路：首先看到这个函数不要被吓到，实际上就是求多少种子树

            dp[u][0]表示以u为根，子树大小为偶数的方案数

            dp[u][1]表示以u为根，子树大小为奇数的方案数

            这样是按升序算的结果，将这个乘2可以得到升序+降序的结果，但是中间会有部分算重

            考虑这样两种情况

            1.如果u的每个孩子大小都为偶数，那么选任意数量的孩子的这种情况都是重复了

            2.如果u的每个孩子大小都为奇数，那么选任意奇数数量的孩子的这种情况也是重复了（自己画一下图就知道了）

            细节看代码~

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

const int MOD = (int)1e9+7;
typedef long long ll;
const int MAXN = 100005;
int n;
vectorhead[MAXN];
ll dp[MAXN][2];
ll dp2[2][2];

ll add(ll x,ll y)
{
    x+=y;
    if(x>=MOD)x-=MOD;
    return x;
}

ll sub(ll x,ll y)
{
    x-=y;
    if(x<0)x+=MOD;
    return x;
}

void dfs(int u)
{
    dp[u][1]=1;
    dp[u][0]=0;
    int L=head[u].size();
    if(!L)return;
    for(int i=0;i