// Slower C++ solution:
#include <bits/stdc++.h>                                           

 

//#define int long long

//#pragma GCC optimize("Ofast")

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4")

 

 

#define file(s) freopen(s".in","r",stdin); freopen(s".out","w",stdout);

#define forev(i, b, a) for(int i = (b); i >= (a); --i)

#define forn(i, a, b) for(int i = (a); i <= (b); ++i)

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

#define sz(s) (int)s.size()

#define pb push_back

#define ppb pop_back

#define s second

#define f first

 

 

using namespace std;

 

 

typedef pair < long long, long long > pll;    

typedef pair < int, int > pii;

typedef unsigned long long ull;         

typedef vector < pii > vpii;

typedef vector < int > vi;

typedef long double ldb;  

typedef long long ll;  

typedef double db;                      

 

 

const int inf = 1e9, maxn = 4e5 + 148, mod = 1e9 + 7, N = 1e5 + 11;

const int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1}, block = 555;

const pii base = {1171, 3307}, Mod = {1e9 + 7, 1e9 + 9};

const db eps = 1e-12, pi = 3.14159265359;

const ll INF = 1e18;





int n, v, u, dp[N][2];

vi g[N];





void dfs (int v, int p = 0) {

	dp[v][0] = 1;

	dp[v][1] = 1;



	for (auto to : g[v])

		if (to != p) {

			dfs(to, v);

			dp[v][0] = dp[v][0] * 1ll * dp[to][1] % mod;

			dp[v][1] = dp[v][1] * 1ll * (dp[to][0] + dp[to][1]) % mod;

		}

}





main () {

	cin >> n;

	forn (i, 1, n - 1) {

		cin >> v >> u;

		g[v].pb(u);

		g[u].pb(v);

	}



	dfs(1);

	cout << (dp[1][0] + dp[1][1]) % mod;

}



    

// Faster C++ solution for the same problem (measured speedup: 3.444465790997766x):
#include <bits/stdc++.h>

#define f first

#define s second

#define fore(i,a,b) for(int i = (a), ThxMK = (b); i < ThxMK; ++i)

#define pb push_back

#define all(s) begin(s), end(s)

#define _ ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define sz(s) int(s.size())

#define ENDL '\n'

using namespace std;

typedef long double ld;

typedef long long lli;

typedef pair<lli,lli> ii;

typedef vector<int> vi;

typedef vector<lli> vl;

typedef vector<ii> vii;

typedef vector<vi> vvi;

typedef vector<vl> vvl;

#define deb(x) cout << #x": " << (x) << endl;



const int N = 3e3+5;

const lli INF = 1e18;

const lli MOD = 1e9+7;



vi m;

vi primes(1,2);

vector<vector<ld>>nk;

void sieve(){

	m = vi(N+1,0);

	for(int i = 4; i<N; i+=2)m[i]=1;

	for(int i = 3; i*i<=N; i+=2)if(!m[i])for(int j = i*i; j<N; j+=i)m[j]=1;

	for(int i = 3; i<N; i+=2)if(!m[i])primes.pb(i);

}



void pascal(){

	nk =vector<vector<ld>>(N,vector<ld>(N,0.0));

	fore(i,0,N)nk[i][0]=nk[i][i]=1.0;

	fore(i,1,N)fore(j,1,i)nk[i][j]=nk[i-1][j-1]+nk[i-1][j];

}



lli gcd(lli a, lli b){return (b?gcd(b,a%b):a);}



lli lcm(lli a, lli b){

	if(a<b)swap(a,b);

	lli c = gcd(a,b);

	a/=c;

	return a*b;

}



int popcount(lli x) { return __builtin_popcountll(x); }



lli poww(lli a, lli b){

	lli res =1;

	while(b){

		if(b&1) res = res * a;

		a = a*a;

		b/=2;

	}

	return res;

}



lli powm(lli a, lli b){

	lli res =1;

	while(b){

		if(b&1) res = (res * a)%MOD;

		a = (a*a)%MOD;

		b/=2;

	}

	return res;

}



bool isPrime(lli x){

	if(x==1)return 0;

	for(auto i:primes){

		if(i*i>x)return 1;

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

	}

	return 1;

}



lli SUM(lli a, lli b, lli mod=MOD){

	return (a+b+mod)%mod;

}

lli PROD(lli a, lli b, lli mod=MOD){

	a%=mod; b%=mod;

	return a*b % mod;

}

int dx[4]={-1,0,0,1}, dy[4]={0,1,-1,0};



// ---- コーディングはここから！ ('-')7



void solve(){

	int n; cin>>n;

	vvi v(n+1);

	fore(i,0,n-1){

		int a,b; cin>>a>>b;

		v[a].pb(b);

		v[b].pb(a);

	}

	if(n==1){cout<<2<<ENDL;return;}

	vvl dp(n+1,vl(2,1));

	auto f=[&](auto f, int u, int pr)->void{

		for(auto i:v[u])if(i!=pr)f(f,i,u);

		for(auto i:v[u])if(i!=pr)dp[u][1]=PROD(dp[u][1],SUM(dp[i][0],dp[i][1]));

		for(auto i:v[u])if(i!=pr)dp[u][0]=PROD(dp[u][0],dp[i][1]);

	};

	f(f,1,1);



	cout<<SUM(dp[1][0],dp[1][1])<<ENDL;

}



int main(){_

	//int t; cin>>t; while(t--)

	solve();

}