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++14 (GCC 5.4.1) (judge verdict: Accepted)

/* Dsingh_24 */

#include<bits/stdc++.h>
#define ll          long long
#define pb          push_back
#define	endl		'\n'
#define pii         pair<ll int,ll int>
#define vi          vector<ll int>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll int)x.size()
#define hell        1000000007
#define rep(i,a,b)	for(ll int i=a;i<b;i++)
#define lbnd        lower_bound
#define ubnd        upper_bound
#define bs          binary_search
#define mp          make_pair
#define ios	    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;

#define N  100005
ll n,x,y;
vi a[N];
ll dp[N][2];
ll dfs(ll node,ll p){
	for(auto i:a[node]){
		if(i!=p){
			//cout<<"node="<<node<<" "<<dp[node][0]<<" "<<dp[node][1]<<"\n";
			dfs(i,node);
			dp[node][0]*=((dp[i][0]+dp[i][1])%hell);
			dp[node][0]%=hell;
			dp[node][1]*=dp[i][0];
			dp[node][1]%=hell;
		}
	}
	if(a[node].size()==1&&p!=0)
		dp[node][0]=1,dp[node][1]=1;
}
int main()
{
	ios
	cin>>n;
	ll i,j;
	if(n==1){cout<<"2";return 0;}
	for(i=0;i<n-1;i++){
		cin>>x>>y;
		a[x].pb(y);
		a[y].pb(x);
	}
	for(i=0;i<N;i++)
		for(j=0;j<2;j++)
			dp[i][j]=1;
	dfs(1,0);
	cout<<(dp[1][0]+dp[1][1])%hell;
	return 0;
}