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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;
import java.io.*;

public class Main {

    static StreamTokenizer in;

    static int nextInt() throws IOException {
        in.nextToken();
        return (int) in.nval;
    }

    static int mod = (int) (1e9 + 7);
    static LinkedList<Integer>[] linkedLists;

    public static void main(String[] args) throws Exception {
        in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

        int n = nextInt();

        linkedLists = new LinkedList[n];

        for (int i = 0; i < n; i++) {
            linkedLists[i] = new LinkedList<>();
        }

        for (int i = 0; i < n-1; i++) {
            int node1 = nextInt() - 1;
            int node2 = nextInt() - 1;

            linkedLists[node1].add(node2);
            linkedLists[node2].add(node1);
        }

        Pair p = dfs(0, -1);

        out.println((p.numBlack + p.numWhite) % mod);

        out.close();
    }

    static int multiply(long a, int b) {

        return (int) ((a * b) % mod);
    }

    static Pair dfs(int node, int parent) {
        int white = 1; // always can be white
        int black = 0; // can't always be black

        for (int i : linkedLists[node]) {
            if (i != parent) {
                Pair p = dfs(i, node);
                int oldWhite = white;
                white = multiply(white, p.numWhite);
                black  = (multiply(black, (p.numWhite + p.numBlack) % mod) + multiply(oldWhite, p.numBlack)) % mod;
            }
        }

        return new Pair((white + black) % mod, white);
    }

    static class Pair {
        int numWhite, numBlack; // p.numWhite is the number of ways to color the subtree such that p is white, and conversely p.numBlack is the number of ways when this is black

        Pair(int numWhite, int numBlack) {
            this.numWhite = numWhite;
            this.numBlack = numBlack;
        }
    }
}