// Original C++ solution (slower):
/*

created by sanyam singhal

*/



#include<bits/stdc++.h>

using namespace std;



#define pb push_back

#define mp make_pair

#define ll long long int

#define mod 1000000007

#define w(t) int t;	cin>>t while(t--)

#define triplet pair<int,pair<int,int>>

#define vi vector<int>

#define vll vector<ll>

#define pi pair<int,int>

#define pll pair<ll,ll>

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

#define loop(i, k, n) for(ll i=k; i<n; i+=1)

#define loop2(i, k, n) for(ll i=k; i>=n; i--)



const ll inf = 1e18;

ll dp[100005][4];

bool vis[100005];



void take_fileIO(){

  #ifndef ONLINE_JUDGE

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

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

  #endif

  return;

}



ll solve_by_dp(vector<vector<int>> &adj, int node, int parent_color){

  if(dp[node][parent_color]!=-1)

    return dp[node][parent_color];



  vis[node] = true;

  ll ans = 1;

  for(int neigh : adj[node]){

    if(vis[neigh]==false){



      if(parent_color==0){  // 0 here for white

        ll op1 = solve_by_dp(adj, neigh, parent_color)%mod;



        ll op2 = solve_by_dp(adj, neigh,1-parent_color)%mod;

      

        ans = (ans%mod * (op1 + op2)%mod)%mod;



        // cout<<"neigh and node are "<<neigh<<" "<<node<<"color "<<parent_color<<" "<<ans<<endl;

      }

      else{ // 1 for black

        ans = (ans%mod * solve_by_dp(adj, neigh, 1-parent_color)%mod)%mod;

        // cout<<"neigh and node are "<<neigh<<" "<<node<<"color "<<parent_color<<" "<<ans<<endl;

      }

      

    }

  }



  vis[node]=false;

  dp[node][parent_color] = ans;

  return ans;

}



void solve(){

  int n;  

  cin>>n;



  vector<vector<int>>adj(n+1);

  loop(i, 0, n-1){

    int u, v;

    cin>>u>>v;

    adj[u].push_back(v);

    adj[v].push_back(u);

  }



  // loop(i, 1, n+1){

  //   cout<<i<<"->";

  //   for(int neigh : adj[i]){

  //     cout<<neigh<<" ";

  //   }

  //   cout<<endl;

  // }



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

  memset(vis, false, sizeof(vis));



  ll ans = solve_by_dp(adj, 1, 0)%mod;

  // cout<<ans<<endl;



  

  // memset(vis, false, sizeof(vis));



  // cout<<"---------------\n";



  ll ans2 = solve_by_dp(adj, 1, 1)%mod;



  // cout<<ans2<<endl;



  ll f_ans = (ans%mod + ans2%mod)%mod;



  cout<<f_ans<<endl;



  return;

}



int main(){



  fastIO;

  // take_fileIO();

  solve();

  return 0;

}

// Optimized solution to the same problem (faster):
/*

  Very Nice question of DP on Trees.



  Here we are required to find the possible ways to paint it with given conditions.

  So what we can do is:- 

  take every node and visit its children.

  If that node was painted black then we are bound to color its childs as white, so just recur for every

  child by giving white color and multiply the result for every child branch



  If that node was painted white then we have two choice for every child to be painted, so in addition to above 

  calculation do the calculation by assuming the childs to be white also.

  Then finally add the above two answers.



  So here dp array is dp[total_number_of_nodes][colors_possible]



  then answer is just sum of both possible colors for root node.



  created by sanyam singhal

*/



#include<bits/stdc++.h>

using namespace std;



#define pb push_back

#define mp make_pair

#define ll long long int

#define mod 1000000007

#define w(t) int t;	cin>>t while(t--)

#define triplet pair<int,pair<int,int>>

#define vi vector<int>

#define vll vector<ll>

#define pi pair<int,int>

#define pll pair<ll,ll>

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

#define loop(i, k, n) for(ll i=k; i<n; i+=1)

#define loop2(i, k, n) for(ll i=k; i>=n; i--)



const ll inf = 1e18;

ll dp[100001][2];

bool vis[100001];



void take_fileIO(){

  #ifndef ONLINE_JUDGE

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

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

  #endif

  return;

}



ll solve_by_dp(vector<vector<int>> &adj, int node, int parent_color){

  if(dp[node][parent_color]!=-1)

    return dp[node][parent_color];



  vis[node] = true;

  ll ans = 1;

  for(int neigh : adj[node]){

    if(vis[neigh]==false){



      if(parent_color==0){  // 0 here for white

        ll op1 = solve_by_dp(adj, neigh, parent_color)%mod;



        ll op2 = solve_by_dp(adj, neigh,1-parent_color)%mod;

      

        ans = (ans%mod * (op1 + op2)%mod)%mod;



        // cout<<"neigh and node are "<<neigh<<" "<<node<<"color "<<parent_color<<" "<<ans<<endl;

      }

      else{ // 1 for black

        ans = (ans%mod * solve_by_dp(adj, neigh, 1-parent_color)%mod)%mod;

        // cout<<"neigh and node are "<<neigh<<" "<<node<<"color "<<parent_color<<" "<<ans<<endl;

      }

      

    }

  }



  vis[node]=false;

  dp[node][parent_color] = ans;

  return ans;

}



void solve(){

  int n;  

  cin>>n;



  vector<vector<int>>adj(n+1);

  loop(i, 0, n-1){

    int u, v;

    cin>>u>>v;

    adj[u].push_back(v);

    adj[v].push_back(u);

  }



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

  memset(vis, false, sizeof(vis));



  ll ans = solve_by_dp(adj, 1, 0)%mod;  //by taking root as white

  ll ans2 = solve_by_dp(adj, 1, 1)%mod; //by taking root as black



  ll f_ans = (ans%mod + ans2%mod)%mod;

  cout<<f_ans<<endl;

  return;

}



int main(){



  fastIO;

  //take_fileIO();

  solve();

  return 0;

}

// Fastest known solution for this problem: