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)

#include <bits/stdc++.h>
//#include <ext/numeric>
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including tree_order_statistics_node_update
#define oo 0x3f3f3f3f
#define OO 0x3f3f3f3f3f3f3f3f
#define ones(n) __builtin_popcount(n)
#define ONES(n) __builtin_popcountll(n)

using namespace std;
//using namespace __gnu_cxx;
using namespace __gnu_pbds;
template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

typedef pair<int, int> ii;
typedef pair<long long, pair<int, int>> iii;
typedef pair<ii, ii> iiii;

const double PI = acos(-1.0), EPS = 1e-6;
const int N = 402, M = 100005, mod = 1000000007, mxLog = 20;
vector<int> adj[N];
int memo[N][2];

int solve(int u, int par, int par_color) {
    if (adj[u].size() == 0 || adj[u].size() == 1 && par != -1) {
        return 1 + par_color;
    }
    int &ret = memo[u][par_color];
    if (~ret)
        return ret;
    ret = 0;
    if (par_color) {
        long long x = 1;
        for (int v : adj[u]) {
            if (v == par)
                continue;
            x *= 1LL * solve(v, u, 1);
            x %= mod;
        }
        ret += x;
        x = 1;
        for (int v : adj[u]) {
            if (v == par)
                continue;
            x *= 1LL * solve(v, u, 0);
            x %= mod;
        }
        ret += x;
        ret %= mod;
    } else {
        long long x = 1;
        for (int v : adj[u]) {
            if (v == par)
                continue;
            x *= 1LL * solve(v, u, 1);
            x %= mod;
        }
        ret += x;
    }
    return ret;
}



int main() {
#ifndef ONLINE_JUDGE
//    freopen("input.txt", "rt", stdin);
//    freopen("output.txt", "w", stdout);
#endif
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0), cout.precision(10), cout << fixed;

    int n;
    cin >> n;
    int u, v;
    for (int i = 1; i < n; ++i) {
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    memset(memo, -1, sizeof memo);
    cout << solve(1, -1, 1) << '\n';


    return 0;
}