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: Accepted)


#include "bits/stdc++.h"
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;

#define ub              upper_bound
#define lb              lower_bound
#define isrt            insert
#define clr             clear
#define rsz             resize
#define ff              first
#define ss              second
#define lli             long long int
#define pb              push_back
#define pf              push_front
#define mkp              make_pair
#define pii             pair<int,int>
#define vi              vector<int>
#define mii             map<lli,lli>
#define pqb             priority_queue<int>
#define pqs             priority_queue<int,vi,greater<int> >
#define setbits(x)      __builtin_popcountll(x)
#define zrobits(x)      __builtin_ctzll(x)
#define mod             1000000007
#define INF             1e9
#define ps(x,y)         fixed<<setprecision(y)<<x
#define mk(arr,n,type)  type *arr=new type[n];
#define w(x)            int x; cin>>x; while(x--)
#define all(v)          v.begin(),v.end()
#define endl            "\n"


typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;

lli maxm(lli a, lli b)
{
    return (a >= b ? a : b);
}


lli minm(lli a, lli b)
{
    return (a <= b ? a : b);
}

lli power(lli x, lli y, lli p)
{
    lli res = 1;

    x = x % p;
    if (x == 0)
        return 0;

    while (y > 0)
    {

        if (y & 1)
            res = (res * x) % p;


        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}

lli nCrmodp(lli n, lli r, lli p)
{
    if (r == 0)
        return 1;
    if (r == 1)
        return n;


    lli pro = 1, k = n;
    while (k > r)
    {
        pro *= k;
        pro %= p;
        --k;
    }

    k = 2;
    while (k <= (n - r))
    {
        pro *= power(k, (p - 2), p);
        pro %= p;
        ++k;
    }

    return pro;
}


string num_str(lli n, lli k)
{
    string str = "";
    lli x = n;
    while (x != 0)
    {
        string s = to_string(x % k);
        str = s + str;
        x /= k;
    }

    return str;
}


lli first_big(lli n, lli k, lli arr[])
{
    lli l = 0;
    lli r = n - 1;

    lli leftGreater = n;

    while (l <= r) {
        lli m = l + (r - l) / 2;

        if (arr[m] > k) {
            leftGreater = m;
            r = m - 1;
        }

        else
            l = m + 1;
    }

    return leftGreater;
}

void c_p_c()
{
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
}

#define MAXN 100005
vector<int>adj[MAXN];
lli arr[MAXN][2];
vector<bool>vis;

lli dfs(int s, int par, bool st)
{
    vis[s] = true;
    int indx = 0;
    if (st == 1)
        indx = 1;

    if (arr[s][indx] == -1)
    {
        if (adj[s].size() == 1 && s != 1)
            arr[s][indx] = 1;

        else
        {
            lli pro = 1;

            if (indx == 1)
                for (auto it : adj[s])
                {
                    if (it != par)
                        pro *= dfs(it, s, false);
                    pro %= mod;
                }

            else
                for (auto it : adj[s])
                {
                    if (it != par)
                        pro *= (dfs(it, s, true) + dfs(it, s, false));
                    pro %= mod;
                }

            arr[s][indx] = pro;

        }
    }
    return arr[s][indx];
}


int main()
{
    c_p_c();

    int n, u, v;
    cin >> n;

    vis.rsz(n + 1, false);
    memset(arr, -1, sizeof(arr));


    for (int i = 1; i < n; ++i)
    {
        cin >> u >> v;
        adj[u].pb(v);
        adj[v].pb(u);
    }

    if (n == 1)
        cout << "2\n";
    else
        cout << (dfs(1, 0, true) + dfs(1, 0, false)) % mod << endl;

    return 0;
}