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: Time Limit Exceeded)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
 
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("avx2,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
using namespace __gnu_pbds; 
using namespace std;
 
#define ld long double
#define pb push_back
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll long long
#define int long long
#define ull unsigned long long 
#define pii pair<int,int>
#define vii vector<int>
#define f first
#define s second
#define endl "\n"
#define all(container) container.begin(), container.end()
#define onlyunique(v) v.erase(unique(all(v)), v.end());
 
#define TRACE
#ifdef AYUSH
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...)
#endif
 
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class T> ostream& operator << (ostream &os, const vector<T> &v) { for (T i : v) os << i << ' '; return os; }
template <class T> ostream& operator << (ostream &os, const set<T> &v) { for (T i : v) os << i << ' '; return os; }
template <class T> ostream& operator << (ostream &os, const unordered_set<T> &v) { for (T i : v) os << i << ' '; return os; }
template <class T> ostream& operator << (ostream &os, const ordered_set<T> &v) { for (T i : v) os << i << ' '; return os; }
template <class T> ostream& operator << (ostream &os, const multiset<T> &v) { for (T i : v) os << i << ' '; return os; }
template <class T, class S> ostream& operator << (ostream &os, const pair<T, S> &v) { os << '{'<<v.first << ' ' << v.second << '}'; return os; }
template <class T, class S> ostream& operator << (ostream &os, const unordered_map<T, S> &v) { for (auto i : v) os << '(' << i.first << "=>" << i.second << ')' << ' '; return os; }
template <class T, class S> ostream& operator << (ostream &os, const map<T, S> &v) { for (auto i : v) os << '(' << i.first << "=>" << i.second << ')' << ' '; return os; }
//order_of_key (k) : Number of items strictly smaller than k .
//find_by_order(k) : K-th element in a set (counting from zero)
 
int const mod = 1e9+7;
int const inf = 1e18;
 
int power(int x, int y, int mod = 2e18){ int ans = 1;x %= mod;while(y){if(y&1)ans = (x * ans) % mod;x = (x * x) % mod;y >>= 1;}return ans;}
int modinv(int a, int m){return power(a, m-2, m);}
const ld PI = 3.1415926535897932384626433832795;

int n, answer;
vector<vector<int>> g;

int ans(int i, int c, int p = -1) {
    int r = 1;
    for(auto x: g[i]) {
        if(x != p){
            if(c != 0) {
                r *= (ans(x, 1, i) + ans(x, 0, i)) % mod;
            }
            else {
                r *= ans(x, 1, i) % mod;
            }
            r %= mod;
        }
    }
    return r;
}

void solve() {
    cin >> n;
    g.assign(n + 1, vector<int>());
    for (int i = 0; i < n - 1; ++i) {
        int x, y; 
        cin >> x >> y;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    answer = ans(1, 0);
    trace(answer);
    answer = (answer + ans(1, 1)) % mod;
    cout << answer << endl;
}
signed main() {
    IOS
    int t = 1, T = 1;
    // cin>>t;
    while (t--) {
        // cout<<"Case #"<<T<<": ";
        solve();
        // ++T;
    }
    return 0;
}