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: Runtime Error)

#pragma GCC optimize "trapv" 
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define rep(i,a,n)     for(int i=a;i<n;i++)
#define rep3(n)        for(int i=0;i<n;i++)
#define inarr(arr,n)   rep(i,a,n)cin>>arr[i]
#define ll             long long int
#define pb             push_back
#define all(v)         v.begin(),v.end()
#define endl           "\n"
#define eb             emplace_back
#define x              first
#define y              second
#define gcd(a,b)       __gcd(a,b)
#define mem1(a)        memset(a,-1,sizeof(a))
#define pres(a,x)      a.find(x)!=a.end()
#define sz(a)          (int)a.size()
#define pii            pair<int,int>
#define IOS            ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define ret            return 0
#define pi             3.1415926536
#define hell           1000000007
#define narak          998244353
#define elasped_time 1.0 * clock() / CLOCKS_PER_SEC
const int inf1=1e9;
const ll inf2=1e18;
const int N=100000;
using namespace std;
using namespace __gnu_pbds;
#define ordered_set tree<pii, null_type,less<pii>, rb_tree_tag,tree_order_statistics_node_update>
int dirs[4][2] ={{1,0},{-1,0},{0,1},{0,-1}};
int dp[N+1][2];
vector<int>g[N+1];
void dfs(int u,int p){
    dp[u][0]++;
    dp[u][1]++;
    for(auto i:g[u]){
        if(i!=p){
           dfs(i,u);
           dp[u][1]=dp[i][0]*dp[u][1]%hell;
           dp[u][0]=(dp[u][0]*(dp[i][0]+dp[i][1]))%hell;
        }
    }
}
int solve(){
   int n;cin>>n;         
   rep3(n-1){
       int a,b;cin>>a>>b;
       g[a].pb(b);
       g[b].pb(a);
   }          
   dfs(1,0);
   cout<<(dp[1][0]+dp[1][1])%hell<<endl;
 ret;
}
int main(){
    IOS;
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    #ifdef cat
    cat();
    #endif
    int t=1;//cin>>t;
    while(t--){
        solve();
    }
    ret;
}