// Slower C++ solution:
//#pragma GCC target ("avx2,sse2")

//#pragma GCC optimization ("Ofast")

//#pragma GCC optimization ("unroll-loops")

#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;



#define ordered_set tree <pair <ll, int>, null_type, less <pair <ll, int> >, rb_tree_tag, tree_order_statistics_node_update>

#define ll long long

#define ull unsigned long long

#define db double

#define pb push_back

#define pf push_front

#define ppb pop_back

#define ppf pop_front

#define fi first

#define se second

#define mp make_pair

#define up_b upper_bound

#define low_b lower_bound

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

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

#define endl "\n"



#define left fsdsdfoisf

#define sum dpsdfioppsf

#define assign xcvjlkdjfio

#define trie fksdfkjkfnjuiv

#define next sidlfjsfkl

#define merge sdfksdkfsldf



#define int long long



using namespace std;



void dout() {

    cerr << endl;

}

template <typename Head, typename... Tail>

void dout(Head H, Tail... T) {

    cerr << H << ' ';

    dout(T...);

}



//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

typedef pair <int, int> pii;



const int N = 1e5 + 7, MOD = 1e9 + 7;

int n, dp[2][N];

vector <int> q[N];



void dfs(int v, int p) {

    int child = 0, A = 1, B = 1;

    for (auto i : q[v]) {

        if (i == p) {

            continue;

        }

        dfs(i, v);

        A = A * (dp[0][i] + dp[1][i]) % MOD;

        B = B * dp[0][i] % MOD;

        child++;

    }

    if (child == 0) {

        dp[0][v] = dp[1][v] = 1;

        return;

    }

    dp[0][v] = A;

    dp[1][v] = B;

}



void solve(int tc) {

    // check for (int i = 0; i < n; j++)

    cin >> n;

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

        int x, y;

        cin >> x >> y;

        q[x].pb(y);

        q[y].pb(x);

    }

    dfs(1, 1);

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

}



signed main() {

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

//    freopen("input.txt", "r", stdin);

//    freopen("output.txt", "w", stdout);

    int tc = 1;

//    cin >> tc;

    for (int i = 0; i < tc; i++) {

        solve(i);

//        cleanup();

    }

}


// Faster C++ solution for the same problem (measured speedup: 3.4646559745161674x):
#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();

}