// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;



const ll MOD = 1000000007;

const ll MAX = 100100;

const double PI = 3.1415926;



#define vll vector<ll>

#define max_heap priority_queue<int>

#define min_heap priority_queue<int , vector<int>, greater<int> >

#define pb push_back

#define all(container) container.begin(), container.end()

#define pll pair<ll,ll>

#define printArr(name,from,to) for(int x = from; x < to; x++) cout << name[x] << " ";

// #define traverse( container, it) for( typeof(container.begin()) it = container.begin(); it!=container.end(); it++)

int n;

vll adj[MAX];

ll dp[MAX][2];



ll f(ll s, ll p, bool c)

{

	if (adj[s].empty())

		return 1;

	if (adj[s].size() == 1 && p != -1)

		return 1;

	if (dp[s][c] != -1)

		return dp[s][c];



	ll temp = 1;

	for(ll x : adj[s])

	{

		if (x == p)

			continue;

		if(c)

			temp *= f(x,s,!c);

		else

			temp *= (f(x,s,c) + f(x,s,!c));

		temp %= MOD;

	}

	dp[s][c] = temp;

	return temp;

}



int main()

{

	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);

	memset(dp,-1,sizeof(dp));

	cin >> n;

	int x,y;

	for(int i = 0; i < n-1; i++)

	{

		cin >>x>>y;

		adj[x].push_back(y);

		adj[y].push_back(x);

	}

	cout << (f(1,-1,0) + f(1,-1,1))%MOD << endl;

	return 0;

}


// Optimized solution to the same problem (faster):
// You either die a hero, or you live long enough to see yourself become the villain.



/**

 *   author:  Blind_is_love

 *   created: 12.10.2019, 21:07

**/



#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>



using namespace __gnu_pbds;

using namespace std;



typedef long long ll;



template <class T> ostream& operator << (ostream &os, const vector<T> &v) { for (T i : v) os << i << ' '; return os; }

template <class T> ostream& operator << (ostream &os, const set<T> &v) { for (T i : v) os << i << ' '; return os; }

template <class T, class S> ostream& operator << (ostream &os, const pair<T, S> &v) { os << v.first << ' ' << v.second; return os; }

template <class T, class S> ostream& operator << (ostream &os, const map<T, S> &v) { for (auto i : v) os << '(' << i.first << "=>" << i.second << ')' << ' '; return os; }

template <class T, class S> ostream& operator << (ostream &os, const unordered_map<T, S> &v) { for (auto i : v) os << '(' << i.first << "=>" << i.second << ')' << ' '; return os; }

 

#ifndef ONLINE_JUDGE

#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)

    template <class Arg1> void __f(const char* name, Arg1&& arg1) { cerr << name << " : " << arg1 << endl; }

    template <class Arg1, class... Args>

    void __f(const char* names, Arg1&& arg1, Args&&... args) {

        const char* sep = strchr(names + 1, ',');

        cerr.write(names, sep - names) << " : " << arg1 << "  ";

        __f(sep + 1, args...);

    }

#else

#define trace(...) 0

#pragma GCC optimize ("O3")

#pragma GCC optimize ("unroll-loops")

#pragma GCC target("avx2,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#define _CRT_SECURE_NO_WARNINGS

#endif // ifndef ONLINE_JUDGE



#define endl "\n"

#define vll vector<int>

#define max_heap priority_queue<int>

#define min_heap priority_queue<int , vector<int>, greater<int> >

#define pb push_back

#define all(container) container.begin(), container.end()

#define pll pair<ll,ll>

#define pii pair<int,int>

#define printArr(name,from,to) for(int x = from; x < to; x++) {cout << name[x] << " ";} cout << endl;

#define fi first

#define se second

#define mp make_pair

#define prime1 304933

#define prime2 15486277 

#define mod2 179424691

#define int ll



typedef tree<ll, null_type, less<ll>, rb_tree_tag,tree_order_statistics_node_update> ordered_set; 



const ll MOD = 1000000007;

const ll MAX = 1e5+10;

const double PI = 3.1415926;

const ll INF = 1e18;

const double EPS = 1e-5;

int n;

vll adj[MAX];

int dp[MAX][2];

bool seen[MAX][2];



int dfs(int s, int p, bool b) {

	int ans[2]{};

	ans[0] = 1;

	if (!b) ans[1] = 1;



	for(int x : adj[s]){

		if (x == p) continue;

		if (seen[x][false]){

			ans[0] = (ans[0] * dp[x][false])%MOD;

			continue;

		}

		if (seen[x][true]){

			ans[1] = (ans[1] * dp[x][true])%MOD;

			continue;

		}

		ans[0] = (ans[0] * dfs(x,s,false))%MOD;

		ans[1] = (ans[1] * dfs(x,s,true))%MOD;

	}

	seen[s][b] = true;

	return dp[s][b] = (ans[0] + ans[1])%MOD;

}



signed main()

{

	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);

	clock_t t1, t2;

	t1 = clock();

	cin >> n;

	for(int i = 0; i < n-1; ++i){

		int x,y;

		cin >> x >> y;

		x--;y--;

		adj[x].pb(y);

		adj[y].pb(x);

	}

	cout << dfs(0,0,0) << endl;

	t2 = clock(); cerr << "time taken: " << (t2-t1)/(long double)CLOCKS_PER_SEC << endl;

	return 0;

}


// Fastest known solution for this problem: