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>
#define ll          long long int
#define lld          long double
#define pb          push_back
#define pii         pair<ll,ll>
#define mi          map<ll,ll>
#define vec         vector<ll>
#define all(a)     (a).begin(),(a).end()
#define F           first
#define S           second
#define mod         1000000007
#define rep(i,a,b)	for(ll  i=a;i<b;i++)
#define repi(i,a,b)	for(ll  i=a;i<=b;i++)
#define per(i,a,b)  for(ll i=a;i>=b;i--)
#define mp          make_pair
#define mit         map<ll,ll>::iterator
#define sit         set<ll>::iterator
#define pit         pair<ll,ll>::iterator
#define tr(container, it) for(__typeof(container.begin()) it = container.begin(); it != container.end(); it++)
using namespace std;
ll xo(ll x, ll y) 
{ 
   return (x | y) & (~x | ~y); 
}  
ll bin_Expo(ll x,ll n)
{
		if(x==0)
		return 0;
    if(n==0)
        return 1;
    else if(n%2 == 0)        //n is even
        return bin_Expo(x*x,n/2);
    else                             //n is odd
        return x*bin_Expo(x*x,(n-1)/2);
}
ll mod_Expo(ll x,ll n,ll M)
{
		if(x==0)
		return 0;
    if(n==0)
        return 1;
    else if(n%2 == 0)        //n is even
        return mod_Expo((x*x)%M,n/2,M);
    else                             //n is odd
        return (x*mod_Expo((x*x)%M,(n-1)/2,M))%M;

}
bool prime_check(ll x)
{
    bool prime = (x >= 2);
    for (ll i = 2; i * i <= x; i++) 
    {
        if (x % i == 0) 
        {
            prime = false;
            break;
        }
    }
    return prime;
}
ll logg(ll base,ll x) {
    return (ll)(log(x) / log(base));
}
vec v[100005]; ll dp1[100005],dp2[100005];
void dfs(ll x, ll p)
{
	ll y,z;
	y=v[x].size();
	rep(i,0,y)
	{
		z=v[x][i];
		if(z!=p)
		{
			dfs(z,x);
			dp1[x] = (dp1[x]*dp2[z])%mod;
			dp2[x] = (dp2[x]*((dp1[z]+dp2[z])%mod))%mod;
		}
	}
}
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int TESTS=1;
//	cin>>TESTS;
    while(TESTS--)
    {
    	ll n,k,i,j,m,h,w;
    	cin>>n;
    	repi(i,1,n-1)
    	{
    		cin>>h>>w;
    		v[h].pb(w);
    		v[w].pb(h);
		}
		repi(i,1,n){
			dp1[i]=1;
			dp2[i]=1;
		}
		dfs(1,0);
		cout<<(dp1[1]+dp2[1])%mod;
	}
	return 0;
}