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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define fo(i,a,b) for(int i= a ; i < b ; ++i)
#define rep(i,n) fo(i,0,n)
#define pln(n) printf("%lld\n",n)
#define sll(n) scanf("%lld",&n)
#define ss(n) scanf("%s",n)
#define vi vector < int >
#define pii pair < int , int >
#define pb push_back
#define mp make_pair
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define dbg(x) { cout<< #x << ": " << (x) << endl; }
#define dbg2(x,y) { cout<< #x << ": " << (x) << " , " << #y << ": " << (y) << endl; }
#define dbg3(x,y,z) { cout<< #x << ": " << (x) << " , " << #y << ": " << (y) <<" , "<< #z <<": "<<(z)<<endl; }
#define all(x) x.begin(),x.end()
#define sz(x) (int)x.size()
#define vll vector<long long>
#define vvi vector<vector<int>>
#define si set<int>
#define tr(c, it) \
for(decltype(c.begin()) it = c.begin(); it!= c.end(); it++)
#define pis pair<int,string>
#define present(c, x) (c.find(x) != c.end())
#define cpresent(c, x) (find(all(c),x) != c.end())
#define mod (int)(1e9 + 7) 
using namespace std;
vector<int> g[(int)1e5 + 5];


int mul(int a, int b){
	return (ll) a * b %mod;
}
pair<int,int> dfs(int i, int parent){
	
	int black = 0; // at least one is black
	int white = 1; // all white
	for(int ch: g[i]){
		if(ch != parent){
			pair<int,int> p = dfs(ch, i);
			int meme = white;
			white = mul(white, p.second);
			black = mul(meme, p.first) + mul(black, (p.first + p.second))%mod;
		}
	}
	return {white, (black + white)%mod};
	
}

int main(){
	int N;
	cin >> N;
	for(int i = 1; i <= N-1; i++){
		int u, v;
		cin >> u >> v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	pair<int,int> ans = dfs(1, -1);
	cout << (ans.first + ans.second)%mod << endl;
}