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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;

// modint: mod 計算を int を扱うように扱える構造体
template<int MOD> struct Fp {
long long val;
constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
if (val < 0) val += MOD;
}
constexpr int getmod() { return MOD; }
constexpr Fp operator - () const noexcept {
return val ? MOD - val : 0;
}
constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; }
constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; }
constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; }
constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; }
constexpr Fp& operator += (const Fp& r) noexcept {
val += r.val;
if (val >= MOD) val -= MOD;
return *this;
}
constexpr Fp& operator -= (const Fp& r) noexcept {
val -= r.val;
if (val < 0) val += MOD;
return *this;
}
constexpr Fp& operator *= (const Fp& r) noexcept {
val = val * r.val % MOD;
return *this;
}
constexpr Fp& operator /= (const Fp& r) noexcept {
long long a = r.val, b = MOD, u = 1, v = 0;
while (b) {
long long t = a / b;
a -= t * b; swap(a, b);
u -= t * v; swap(u, v);
}
val = val * u % MOD;
if (val < 0) val += MOD;
return *this;
}
constexpr bool operator == (const Fp& r) const noexcept {
return this->val == r.val;
}
constexpr bool operator != (const Fp& r) const noexcept {
return this->val != r.val;
}
friend constexpr ostream& operator << (ostream &os, const Fp<MOD>& x) noexcept {
return os << x.val;
}
friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
if (n == 0) return 1;
auto t = modpow(a, n / 2);
t = t * t;
if (n & 1) t = t * a;
return t;
}
};

const int MOD = 1000000007;
using mint = Fp<MOD>;

const int MAX_N = 100050;

vector<int> G[MAX_N];

struct DP{
    mint white;
    mint black;
    DP() : white(1),black(1){}
    DP &operator+=(const DP &other) {
        white *= other.white + other.black;
        black *= other.white;
        return *this;
    }
};

DP dfs(int v = 0, int p = -1) {
    DP dp;
    for(int nv : G[v]) {
        if(nv == p) continue;
        dp += dfs(nv,v);
    }
    return dp;
}

int main() {
int N;
cin >> N;
rep(i,N-1) {
    int x,y;
    cin >> x >> y;
    x--; y--;
    G[x].push_back(y);
    G[y].push_back(x);
}
DP ans = dfs(0);
cout << ans.black + ans.white << endl;
return 0;
}