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

using namespace std;

using ll = long long;

using P = pair<ll,ll>;

using Map = map<string,ll>;

using vl = vector<ll>;

using vvl = vector<vector<ll>>;

ll INF=1LL<<60;

ll MOD=1000000007;

ll N;

vector<vector<ll>> graph(100005, vector<ll>(0));



void dfs (ll x, vl &check, vvl &children){

    check[x]=1;

    for(ll i=0;i<graph[x].size();i++){

        if(check[graph[x][i]]==0){

            children[x].emplace_back(graph[x][i]);

            dfs(graph[x][i],check,children);

        }

    ;}

}



vector<ll> dpb(100005,-1);

vector<ll> dpw(100005,-1);



ll answ(ll x, vvl &children);



ll ansb(ll x, vvl &children){

    if(dpb[x]!=-1)return dpb[x];

    ll s=1;

    for(ll i=0;i<children[x].size();i++){

        s*=answ(children[x][i],children);

        s%=MOD;

    ;}

    return dpb[x]=s;

}

ll answ(ll x, vvl &children){

    if(dpw[x]!=-1)return dpw[x];

    ll s=1;

    for(ll i=0;i<children[x].size();i++){

        s*=answ(children[x][i],children)+ansb(children[x][i],children);

        s%=MOD;

    ;}

    return dpw[x]=s;

}



int main(){

    cin>>N;

    vector<ll> x(N-1,0);

    vector<ll> y(N-1,0);

    for(ll i=0;i<N-1;i++){

        cin>>x[i]>>y[i]

    ;}

    for(ll i=0;i<N-1;i++){

        graph[x[i]].emplace_back(y[i]);

        graph[y[i]].emplace_back(x[i]);

    ;}

    vector<vector<ll>> graph(N+1, vector<ll>(0));



    //1を根にします！



    vector<ll> check(N+1,0);

    vector<vector<ll>> children(N+1, vector<ll>(0));

    dfs(1,check,children);

    cout<<(ansb(1,children)+answ(1,children))%MOD<<endl;







    return 0;

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

using ll = long long;

using ld = long double;

using P = pair<ll,ll>;

using vl = vector<ll>;

using Map = map<ll,ll>;

using Tup = tuple<ll,ll,ll>;

using vvl = vector<vector<ll>>;

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

#define prt(v) cout<<(v)<<"\n";

#define fl cout<<flush;

#define fi(v) get<0>(v)

#define se(v) get<1>(v)

#define th(v) get<2>(v)

#define endl "\n"

template <typename T> inline bool chmax(T &a, const T &b){if (a<b){a=b;return 1;}return 0;}

template <typename T> inline bool chmin(T &a, const T &b){if (a>b){a=b;return 1;}return 0;}

const ll INF=1LL<<60;

const ll MOD=1000000007;

const ll MOD2=998244353;

const ld pi=3.141592653589793238;



//1-indexedなので注意！！！！

//＄１には頂点数＋３くらいをいれるといい

vector<vector<ll>> graph(100003, vector<ll>(0));  // DAG

vector<ll> order(0);  // トポロジカルソートされた頂点を格納

vector<bool> used(100003,0);  // 訪問済みフラグ



// 辿れるところまで辿る深さ優先探索

void dfs(ll u){

    if(used[u]) return;

    used[u] = true;

    for(ll v : graph[u]) dfs(v);

    order.emplace_back(u);     // 帰りがけ順で頂点を入れていく

}



// トポロジカルソート

//Nは頂点数

void tsort(ll N){

    for(ll v = 1; v <= N; v++) dfs(v);

    //reverse(order.begin(), order.end());

}

//1-indexedなので注意！！！！

//1-indexedなので注意！！！！

//1-indexedなので注意！！！！

//1-indexedなので注意！！！！



signed main(void){

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

    ll N;

    cin >> N;

    vector<ll> x(N-1,0);

    vector<ll> y(N-1,0);

    for(ll i=0;i<N-1;++i){

        cin>>x[i]>>y[i];

        graph[x[i]].emplace_back(y[i]);

        graph[y[i]].emplace_back(x[i]);

    }

    tsort(N);

    vector<bool> check(N+1,0);

    vector<ll> dpb(N+1,0);

    vector<ll> dpw(N+1,0);

    for(auto p: order){

        check[p]=1;

        ll retw=1;

        ll retb=1;

        for(auto q: graph[p]){

            if(!check[q])continue;

            retw*=dpw[q]+dpb[q];

            retb*=dpw[q];

            retw%=MOD;

            retb%=MOD;

        }

        dpw[p]=retw;

        dpb[p]=retb;

    }

    prt((dpw[order[N-1]]+dpb[order[N-1]])%MOD)





    return 0;

}


// Fastest known solution for this problem: