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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
    
#define IOS cin.sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define rep(i, n) for (int i = 0; i < n; i++)
#define repn(i, a, b) for (int i = a;i<=b;i++)
#define ll long long int
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define mem(x) memset(x, 0, sizeof(x))
#define ritr(it, a) for (auto it = a.begin(); it != a.end(); it++)
#define PI 3.1415926535897932384626
#define pai pair < int, int> 
#define pal pair < ll, ll> 
#define vi vector <int> 
#define vl vector < ll > 
#define vpai vector < pii > 

const int mod = 1e9+7;
const int INF = INT_MAX;
const int sze = 100005;

ll dp[sze][2];
vi adj[sze];

void dfs(int start,int par){
    if(adj[start].size()==0){
        dp[start][0]=1;
        dp[start][1]=1;
        return;
    }
    ll a=1,b=1;
    for(auto x:adj[start]){
        if(par==x) continue;
        dfs(x,start);
        a*=(dp[x][0]+dp[x][1])%mod;
        b*=(dp[x][0])%mod;
    }
    dp[start][0]=a;
    dp[start][1]=b;
}

void solve() {
    int n;
    cin>>n;
    int m=n-1;
    while(m--){
        int a,b;
        cin>>a>>b;
        a--;
        b--;
        adj[a].pb(b);
        adj[b].pb(a);
    }
    dfs(0,-1);
    cout<<dp[0][0]+dp[0][1]<<endl;
}

int main() {
    IOS;

    int t = 1;
    // cin >> t;
    while (t--) {
        mem(dp);
       solve();
    }

  return 0;
}