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: Wrong Answer)

//----------------------------templates
    #pragma GCC optimize ("Ofast")
    #pragma GCC target ("tune=native")
    #pragma GCC target ("avx")
    //----------------------------
    #include <bits/stdc++.h>
    using namespace std;

    typedef long long ll;
    typedef unsigned long long ull;
    #define int ll

    #define FOR(i,j,n) for (int i=(int)(j);i<(n);i++)
    #define REP(i,n) for (int i=0;i<(int)(n);i++)
    #define REPS(i,n) for (int i=1;i<=(int)(n);i++)
    #define REPN(i,n) for (int i=(int)(n)-1;i>=0;i--)
    #define REPNS(i,n) for (int i=(int)(n);i>0;i--)

    #define I(n) scanf("%lld", &(n))
    #define LL(n) scanf("%lld", &(n))
    #define pb(n) push_back((n))
    #define mp(i,j) make_pair((i),(j))
    #define eb(i,j) emplace_back((i),(j))
    #define y0 y3487465
    #define y1 y8687969
    #define j0 j1347829
    #define j1 j234892
    #define uniq(v) v.erase( unique(v.begin(), v.end()), v.end() )

    #define all(x) (x).begin(),(x).end()
    #define sz(x) ((int)(x).size())

    typedef vector<int> vi;
    typedef pair<int,int> pi;
    typedef vector<pi> vpi;
    typedef vector<vi> vvi;
    typedef vector<vpi> vvpi;
    typedef vector<vvi> vvvi;

    const int mod = 1000000007;

//--------------------------------------------

int n,x,y;
vvi edges;
deque<int> q;
int dp[2][100010],visited[100010], ret;

int accum(int n){
    visited[n] = 1;
    dp[0][n] = 1;
    dp[1][n] = 1;
    for(auto ch: edges[n]){
        if (visited[ch]) continue;
        accum(ch);
        dp[0][n] *= (dp[0][ch] + dp[1][ch]);
        dp[0][n] %= mod;
        dp[1][n] *= (dp[0][ch]);
        dp[1][n] %= mod;
    }
    return dp[0][n] + dp[1][n];
}

signed main(){
    I(n);
    edges.resize(n);
    REP(i,n-1){
        I(x); I(y);
        x--; y--;
        edges[x].pb(y);
        edges[y].pb(x);
    }
    
    cout << accum(0) << endl;
}