#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

#define N 100001

#define INF ~(1 << 31)

using namespace std;

int n, K, cnt, root, tot, ans;

int head[N], to[N], nex[N], val[N], f[N], size[N], deep[N], d[N];

bool vis[N];

//f[i]表示节点i的最大子树的大小

//deep[i]表示节点i到根的距离

//vis[i] == 1 表示该节点删除 

inline int read()

{

	int x = 0, f = 1;

	char ch = getchar();

	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;

	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';

	return x * f;

}

inline void add(int x, int y, int z)

{

	to[cnt] = y;

	val[cnt] = z;

	nex[cnt] = head[x];

	head[x] = cnt++;

}

//获取当前块的重心

inline void getroot(int u, int fa)

{

	int i, v;

	f[u] = 0;

	size[u] = 1;

	for(i = head[u]; ~i; i = nex[i])

	{

		v = to[i];

		if(v == fa || vis[v]) continue;

		getroot(v, u);

		size[u] += size[v];

		f[u] = max(f[u], size[v]);

	}

	f[u] = max(f[u], tot - size[u]);

	if(f[u] < f[root]) root = u;

}

//获取当前块中所有的点到根的距离

inline void getdeep(int u, int fa)

{

	int i, v;

	deep[++deep[0]] = d[u];

	for(i = head[u]; ~i; i = nex[i])

	{

		v = to[i];

		if(v == fa || vis[v]) continue;

		d[v] = d[u] + val[i];

		getdeep(v, u);

	}

}

//求解经过当前根的满足条件的路径的方案数

inline int cal(int u, int now)

{

	int l, r, ret = 0;

	d[u] = now;

	deep[0] = 0;

	getdeep(u, 0);

	sort(deep + 1, deep + deep[0] + 1);

	l = 1, r = deep[0];

	while(l < r)

		if(deep[l] + deep[r] <= K) ret += r - l++;

		else r--;

	return ret;

}

//递归求解每一块

inline void work(int u)

{

	int i, v;

	vis[u] = 1;

	ans += cal(u, 0);

	for(i = head[u]; ~i; i = nex[i])

	{

		v = to[i];

		if(vis[v]) continue;

		//因为在求解的时候会遇到这种情况：经过当前根的满足条件的路径的两端点在同一颗子树中，这样的路径也会统计到答案中

		//然而并不合法，所以需要遍历每一颗子树，减去每一颗子树中满足条件的路径

		ans -= cal(v, val[i]);

		tot = size[v];

		root = 0;

		getroot(v, u);

		work(root);

	}

}

int main()

{

	int i, x, y, z;

	while(~scanf("%d %d", &n, &K))

	{

		if(!n && !K) break;

		cnt = root = ans = 0;

		memset(vis, 0, sizeof(vis));

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

		for(i = 1; i < n; i++)

		{

			x = read();

			y = read();

			z = read();

			add(x, y, z);

			add(y, x, z);

		}

		tot = n;

		f[0] = INF;

		getroot(1, 0);

		work(root);

		printf("%d\n", ans);

	}

	return 0;

}