Problem:
Score : 100 points

Problem Statement

There is a tree with N vertices, numbered 1, 2, \ldots, N.
For each i (1 \leq i \leq N - 1), the i-th edge connects Vertex x_i and y_i.

Taro has decided to paint each vertex in white or black.
Here, it is not allowed to paint two adjacent vertices both in black.

Find the number of ways in which the vertices can be painted, modulo 10^9 + 7.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq x_i, y_i \leq N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
x_2 y_2
:
x_{N - 1} y_{N - 1}

Output

Print the number of ways in which the vertices can be painted, modulo 10^9 + 7.

Sample Input 1

3
1 2
2 3

Sample Output 1

5

There are five ways to paint the vertices, as follows:

Sample Input 2

4
1 2
1 3
1 4

Sample Output 2

9

There are nine ways to paint the vertices, as follows:

Sample Input 3

1

Sample Output 3

2

Sample Input 4

10
8 5
10 8
6 5
1 5
4 8
2 10
3 6
9 2
1 7

Sample Output 4

157

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using static System.Console;
using System.Text;
using System.IO;
using static System.Math;
namespace AtCoder
{
    class Program
    {
        static public int[] Sarray() { return ReadLine().Split().Select(int.Parse).ToArray(); }
        static public List<int> Slist() { return ReadLine().Split().Select(int.Parse).ToList(); }

        static int N;
        static List<int>[] Tree;
        static long[,,] dp;
        static readonly long Mod = 1000000000 + 7;
        static void Main(string[] args)
        {
            N = Sarray()[0];
            Tree = new int[N].Select(_ => new List<int>()).ToArray();
            for (var i = 0; i < N - 1; ++i)
            {
                var xy = Sarray();
                Tree[xy[0] - 1].Add(xy[1] - 1);
                Tree[xy[1] - 1].Add(xy[0] - 1);
            }

            dp = new long[N, 2, 2];
            for (var i = 0; i < N; ++i)
                for (var j = 0; j < 2; ++j)
                    for (var k = 0; k < 2; ++k)
                        dp[i, j, k] = -1;
            var r = 0;
            WriteLine(calckazu(r, -1, 1, 1));

        }

        //dp[i]=\prod(子の数) iが白の場合の数 　+prod(子の数) iが黒の場合の数 
        //dp[i,sirokazu,kurokazu]=sirokazu\prod dp[i+1,kurokazu+sirokazu,sirokazu] +kurokazu*\prod dp[i+1,sirokazu,0]
        //if(葉)return sirokazu+kurokazu
        static long calckazu(int vno, int preno, int wnum, int bnum)
        {
            if (dp[vno, wnum, bnum] != -1)
                return dp[vno, wnum, bnum];
            long wclc = 1;
            long bclc = 1;
            foreach (var nv in Tree[vno])
            {
                if (nv != preno)
                {
                    if (wnum != 0)
                        wclc *= calckazu(nv, vno, 1, 1);
                    if (bnum != 0)
                        bclc *= calckazu(nv, vno, 1, 0);
                    wclc %= Mod;
                    bclc %= Mod;
                }
            }
            return dp[vno, wnum, bnum] = ((wnum * wclc) + (bnum * bclc)) % Mod;
        }
    }
}