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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize ("-ffloat-store") // to restrict undesirable precision
#pragma GCC optimize ("-fno-defer-pop")// to pop argument of function as soon as it returns
#define all(a) a.begin(),a.end()
#define ll long long int
#define ld long double
ll power(ll a,ll b,ll m){ if(b==0) return 1; if(b==1) return a%m; ll t=power(a,b/2,m)%m; t=(t*t)%m; if(b&1) t=((t%m)*(a%m))%m; return t;}
ll modInverse(ll a, ll m) { return power(a, m-2, m); }
#define ps push_back
#define fs first
#define sc second
#define takeline cin.ignore();
#define iactive cout.flush();
#define N 100005
#define endl "\n"
#define mod 1000000007
#define PI 3.141592653589793
//((1.0l)*BIG MULTIPLY MAGIC?)
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//-----------------------------------------------
ll n,ar[N][2],dp[N][3]; vector<ll> vc[N];
ll maxval(ll i,ll cl,ll p){
	// cout<<i<<" "<<p<<" "<<cl<<endl;
	if(dp[i][cl]!=-1){
		return dp[i][cl];
	}
	ll an=1,fl=0;
	if(cl==1){
		ll a=1,b=1;
		for(auto it:vc[i]){
			if(it==p) continue;
			a=(a*maxval(it,1,i))%mod;
		}
		for(auto it:vc[i]){
			if(it==p) continue;
			b=(b*maxval(it,2,i))%mod;
		}
		return dp[i][cl]=(a+b)%mod;
	}
	else{
		ll a=1;
		for(auto it:vc[i]){
			if(it==p) continue;
			a=(a*maxval(it,1,i))%mod;
		}
		return dp[i][cl]=a;
	}
}
int main(){

	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

	ll i,j,k,l;
	cin>>n;
	for(i=1;i<n;i++){
		cin>>ar[i][0]>>ar[i][1];
		vc[ar[i][0]].ps(ar[i][1]);
		vc[ar[i][1]].ps(ar[i][0]);
	}
	memset(dp,-1,sizeof(dp));
	ll an=maxval(1,1,0);
	cout<<an;
	return 0;	
}