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)

/*
 _____ _             _              _           _ 
|_   _| |__   ___   / \   _ __  ___| |__  _   _| |
  | | | '_ \ / _ \ / _ \ | '_ \/ __| '_ \| | | | |
  | | | | | |  __// ___ \| | | \__ \ | | | |_| | |
  |_| |_| |_|\___/_/   \_\_| |_|___/_| |_|\__,_|_|                                                

*/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp>
#define ll          long long
#define pb          push_back
#define ppb         pop_back
#define	endl		'\n'
#define mii         map<ll,ll>
#define msi         map<string,ll>
#define mis         map<ll, string>
#define rep(i,a,b)    for(ll i=a;i<b;i++)
#define repr(i,a,b) for(ll i=b-1;i>=a;i--)
#define trav(a, x) 	for(auto& a : x)
#define pii         pair<ll,ll>
#define vi          vector<ll>
#define vii         vector<pair<ll, ll>>
#define vs          vector<string>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll)x.size()
#define hell        1000000007
#define lbnd        lower_bound
#define ubnd        upper_bound
#define max(a,b)	(a>b?a:b)
#define min(a,b)	(a<b?a:b)

/*	For Debugging	*/
#define DEBUG 		cerr<<"\n>>>I'm Here<<<\n"<<endl;
#define display(x) trav(a,x) cout<<a<<" ";cout<<endl;
#define what_is(x)  cerr << #x << " is " << x << endl;

std::mt19937_64 rng(std::chrono::steady_clock::now().time_since_epoch().count());
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update> 
#define TIME        cerr << "\nTime elapsed: " << setprecision(5) <<1000.0 * clock() / CLOCKS_PER_SEC << "ms\n";
#define DECIMAL(n)  cout << fixed ; cout << setprecision(n);
#define FAST ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace __gnu_pbds;
using namespace std;
#define PI 3.141592653589793
#define N  100005
ll dp[N][2];
vi a[N];
ll fun(ll node,bool f,ll pr)
{
	if(dp[node][f]!=-1)
		return dp[node][f];
	ll ans=1;
	for(auto i:a[node])
	{
		if(i!=pr)
		{
			if(f)
			{
				ans=(ans*((fun(i,0,node)+fun(i,1,node))%hell))%hell;
			}
			else
			{
				ans=(ans*((fun(i,1,node))%hell))%hell;
			}
		}
	}
	dp[node][f]=ans;
	return ans;
}
void solve()
{
	ll n;
	cin>>n;
	ll x,y;
	rep(i,1,n)
	{
		cin>>x>>y;
		a[x].pb(y);
		a[y].pb(x);
	}
	rep(i,1,n+1)
	{
		rep(j,0,2)
		dp[i][j]=-1;
	}
	cout<<(fun(1,0,-1)+fun(1,1,-1))%hell;
	return;
}
int main()
{
	FAST
	int TESTS=1;
//	cin>>TESTS;
	while(TESTS--)
	{
		solve();
	}
	TIME
	return 0;
}