#include<iostream>

#include<map>

#include<string>

#include<cstring>

#include<vector>

#include<algorithm>

#include<set>

#include<sstream>

#include<cstdio>

#include<cmath>

#include<climits>

#define endl '\n'

#define ll long long

#define ull unsigned long long

#define fi first

#define se second

#define mp make_pair

#define pii pair<int,int>

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

#define IO ios::sync_with_stdio(false)

#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)

#define per(ii,a,b) for(int ii=b;ii>=a;--ii)

#define forn(x,i) for(int i=head[x];i;i=e[i].next)

#define show(x) cout<<#x<<"="<<x<<endl

#define showa(a,b) cout<<#a<<'['<<b<<"]="baidu<a[b]<<endl

#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl

#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl

#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl

using namespace std;

const int maxn=1e6+10,maxm=2e6+10;

const int INF=0x3f3f3f3f;

const int mod=1e9+7;

const double PI=acos(-1.0);

//head

int casn,n,m,k;

ll val[maxn],dis[maxn],ans,maxt,dfn;

int deep[maxn],vis[maxn],size[maxn];

int dp[maxn],allnode;

struct node {int to,next;ll cost;}e[maxm];int head[maxn],nume;//静态链表存图

void add(int a,int b,ll c){e[++nume]=(node){b,head[a],c};head[a]=nume;}

int mid;

void init(){//初始化

	memset(head,0,sizeof head);

	memset(dis,0,sizeof dis);

	memset(vis,0,sizeof vis);

	nume=0;

}

void getmid(int now,int pre){//dfs求树的重心

	size[now]=1;//当前点为根,其子树的节点数

	for(int i=head[now];i;i=e[i].next){

		int to=e[i].to;

		if(to==pre||vis[to]) continue;

		getmid(to,now);//递归计算子树的大小

		size[now]+=size[to];

	}

	dp[now]=max(size[now],allnode-size[now]);//dp[i]表示以i为根建立子树的时候,最大的子树大小

	if(maxt>dp[now]){//maxt为最大的子树大小

		maxt=dp[now];

		mid=now;

	}

}

void dfs(int now,int pre){//计算深度

	deep[++dfn]=dis[now];

	for(int i=head[now];i;i=e[i].next){

		int to=e[i].to,cost=e[i].cost;

		if(to==pre||vis[to]) continue;

		dis[to]=dis[now]+cost;

		dfs(to,now);

	}

}

int cal(int rt,int len){//计算rt为根的子树中,深度之和>=k的点对数量

	dis[rt]=len,dfn=0;

	dfs(rt,0);//以rt为根,dfs计算其子树中所有点的深度

	sort(deep+1,deep+dfn+1);

	int res=0;

	for(int l=1,r=dfn;l<r;){//排序后从两端向中间逼近,总复杂度nlogn

		if(deep[l]+deep[r]<=k){

			res+=r-l;

			l++;

		}else r--;

	}

	return res;

}

void dc(int rt){//分治以rt为根的子树

	vis[rt]=1;

	ans+=cal(rt,0);//初步计算以rt为根子树答案,包含重复情况

	for(int i=head[rt];i;i=e[i].next){

		int to=e[i].to;

		if(vis[to]) continue;

		ans-=cal(to,e[i].cost);//以子节点为根的子树,设置其距离下界为len,对于其子树而言,如果距离减少子树到rt的距离,依然成立的话,必然会被重复计算

		allnode=size[to];

		mid=0,maxt=INF;

		getmid(to,rt);//寻找以rt为根的子树的重心

		dc(mid);//以子树重心为树上点分治的起点,保证总复杂度为n(logn)^2级别

	}

}

int main() {

//#define test

#ifdef test

	auto _start = chrono::high_resolution_clock::now();

	freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);

#endif

	IO;

	while(cin>>n>>k,n+k){

		init();

		int a,b,c;

		rep(i,2,n){

			cin>>a>>b>>c;

			add(a,b,c);

			add(b,a,c);

		}

		mid=ans=0;

		allnode=n,maxt=INF;

		getmid(1,0);

		dc(mid);

		cout<<ans<<endl;

	}

#ifdef test

	auto _end = chrono::high_resolution_clock::now();

  cerr << "elapsed time: " << chrono::duration<double, milli>(_end - _start).count() << " ms\n";

	fclose(stdin);fclose(stdout);system("out.txt");

#endif

	return 0;

}