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)

#include <bits/stdc++.h>

//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
//using namespace __gnu_pbds;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair <int,int> pii;
typedef pair <ll,ll> pll;

#define si(a)           scanf("%d",&a)
#define sii(a,b)        scanf("%d %d",&a,&b)
#define siii(a,b,c)     scanf("%d %d %d",&a,&b,&c)

#define sl(a)           scanf("%lld",&a)
#define sll(a,b)        scanf("%lld %lld",&a,&b)
#define slll(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)

#define un(x)           x.erase(unique(all(x)), x.end())
#define xx              first
#define yy              second
#define pb              push_back
#define mp              make_pair
#define all(v)          v.begin(),v.end()
#define D(x)            cerr << #x " = " << x << '\n'
#define DBG             cerr << "Hi!" << '\n'

#define CLR(a)          memset(a,0,sizeof(a))
#define SET(a)          memset(a,-1,sizeof(a))

#define PI              acos(-1.0)

//inline int setBit(int n,int pos) { return n = n | (1 << pos); }
//inline int resetBit(int n,int pos) { return n = n & ~(1 << pos); }
//inline bool checkBit(int n,int pos) { return (bool)(n & (1 << pos)); }
//inline int countBit(ll n) { return __builtin_popcountll(n); }


//int fx[] = {+0, +0, +1, -1, -1, +1, -1, +1};
//int fy[] = {-1, +1, +0, +0, +1, +1, -1, -1}; //Four & Eight Direction

/******************************************************************************************/

const int MAX = 100010;
const int INF = 0x3f3f3f3f;
//const double inf = 1.0/0.0;
const int MOD = 1000000007;

vector <int> ed[MAX];
vector <int> G[MAX];

void dfs(int s,int p) {
    for(auto x : ed[s]) {
        if(x == p) continue;
        G[s].pb(x);
        dfs(x,s);
    }
}

int dp[MAX][2];

int call(int s,int p) {

    if(dp[s][p]!=-1) return dp[s][p];

    int ans0 = 1;
    int ans1 = 1;

    for(auto x : G[s]) {
        ans0 *= call(x,0);
        ans0 %= MOD;
        ans1 *= call(x,1);
        ans1 %= MOD;
    }

    if(p==0) dp[s][p] = (ans0 + ans1) % MOD;
    else dp[s][p] = ans0;

    return dp[s][p];
}



int main() {
//    freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

    int n,u,v;
    si(n);

    for(int i=1;i<n;i++) {
        sii(u,v);
        ed[u].pb(v);
        ed[v].pb(u);
    }

    dfs(1,-1);

    SET(dp);

    cout << call(1,0) << endl;

    return 0;
}