Problem:
Score : 100 points

Problem Statement

There is a tree with N vertices, numbered 1, 2, \ldots, N.
For each i (1 \leq i \leq N - 1), the i-th edge connects Vertex x_i and y_i.

Taro has decided to paint each vertex in white or black.
Here, it is not allowed to paint two adjacent vertices both in black.

Find the number of ways in which the vertices can be painted, modulo 10^9 + 7.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq x_i, y_i \leq N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
x_2 y_2
:
x_{N - 1} y_{N - 1}

Output

Print the number of ways in which the vertices can be painted, modulo 10^9 + 7.

Sample Input 1

3
1 2
2 3

Sample Output 1

5

There are five ways to paint the vertices, as follows:

Sample Input 2

4
1 2
1 3
1 4

Sample Output 2

9

There are nine ways to paint the vertices, as follows:

Sample Input 3

1

Sample Output 3

2

Sample Input 4

10
8 5
10 8
6 5
1 5
4 8
2 10
3 6
9 2
1 7

Sample Output 4

157

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define INF ((1LL<<62)-(1LL<<31))
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
#define bit(n,m)(((n)>>(m))&1)
int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}
ll pom(ll a,ll n,int m){ll x=1;for(a%=m;n;n/=2)n&1?x=x*a%m:0,a=a*a%m;return x;}
//#define MOD 998244353
#define MOD 1000000007
#define invp(a,p)pom(a,p-2,p)



//辺の情報を個別に持つタイプ
typedef struct edge{ll s,g;}E;
typedef struct graph{
	ll vcnt,ecnt;
	E  e[200010];//適宜変える
	ll id[100010];//適宜変える
}G;

int esort(const void*a,const void*b){
	E*p=(E*)a,*q=(E*)b;
	if((*p).s<(*q).s)return -1;
	if((*p).s>(*q).s)return  1;
	if((*p).g<(*q).g)return -1;
	return 1;
}
G g;
void readgraph(){
	//適宜変える
	ll n;
	scanf("%lld",&n);
	rep(i,0,n-1){
		ll x,y;
		scanf("%lld%lld",&x,&y);
		x--,y--;
		g.e[2*i].s=x;
		g.e[2*i].g=y;
		g.e[2*i+1].s=y;
		g.e[2*i+1].g=x;
	}
	g.vcnt=n;
	g.ecnt=2*n-2;
	qsort(g.e,g.ecnt,sizeof(E),esort);

	int p=0;
	rep(i,0,g.vcnt){
		while(p<g.ecnt&&g.e[p].s<i)p++;
		g.id[i]=p;
	}
	g.id[g.vcnt]=g.ecnt;//一応番兵
}


//根付き木を下へ塗る
ll dp[100010][2];//dp[i][j]=iを(j?白く:黒く)塗るとき、iを親とする部分木の塗り方
int flag[100010];
int root[100010];

void dfs(int k){
	rep(i,g.id[k],g.id[k+1])if(g.e[i].g!=root[k]){
		root[g.e[i].g]=k;
		dfs(g.e[i].g);
	}
}

void calc(int k){
	if(flag[k])return;
	flag[k]=1;
	dp[k][0]=1;
	dp[k][1]=1;
	//dp[k][0]は子の塗り方の総積
	//dp[k][1]は子のdp[*][0]の総積
	rep(i,g.id[k],g.id[k+1])if(g.e[i].g!=root[k]){
		calc(g.e[i].g);
		dp[k][0]=dp[k][0]*(dp[g.e[i].g][0]+dp[g.e[i].g][1])%MOD;
		dp[k][1]=dp[k][1]*dp[g.e[i].g][0]%MOD;
	}
}

int main(){
	readgraph();
	rep(i,0,g.vcnt)root[i]=-1;
	dfs(0);
	calc(0);
	printf("%lld\n",(dp[0][0]+dp[0][1])%MOD);
}