代码很长 大致思路如下：

类似于CF1009F的长链剖分做法 考虑用长链剖分维护一下某个点的dp数组 然后把询问离线下来 每次做该点的dp数组的前缀和 用的是vector版

#include <bits/stdc++.h>
#define int long long
using namespace std;

int n,q,head[300005],pnt[600005],nxt[600005],E = 0;
int dep[300005],siz[300005],son[300005],fa[300005],d[300005],ans[300005],lazy[300005];
std::vector<int> dp[300005];
std::vector<pair<int,int> > que[300005];
struct Question
{
	int p,k,id;
}a[300005];

bool cmp1(Question a,Question b)
{
	return a.k < b.k;
}

bool cmp2(Question a,Question b)
{
	return a.id < b.id;
}

void add_edge(int u,int v)
{
	pnt[E] = v;
	nxt[E] = head[u];
	head[u] = E++;
}

void dfs1(int u,int f)
{
	d[u] = dep[f] + 1;
	dep[u] = dep[f] + 1;
	son[u] = -1;
	fa[u] = f;
	siz[u] = 1;
	for (int i = head[u]; i != -1; i = nxt[i])
	{
		int v = pnt[i];
		if (v == for) continue;
		dfs1(v,u);
		siz[u] += siz[v];
		d[u] = max(d[u],d[v]);
		if (son[u] == -1 || d[v] > d[son[u]]) son[u] = v;
	}
}

void dfs2(int u,int tp)
{
	if (son[u] == -1)
	{
		dp[u].push_back(0);
		for (int i = 0; i < que[u].size(); i++)
		{
			ans[que[u][i].second] = 0;
		}
		return ;
	}
	dfs2(son[u],tp);
	dp[u].swap(dp[son[u]]);
	for (int i = head[u]; i != -1; i = nxt[i])
	{
		int v = pnt[i];
		if (v == son[u] || v == fa[u]) continue;
		dfs2(v,v);
		for (int j = dp[v].size() - 1; j >= 0; j--)
		{
			dp[u][dp[u].size() - j - 1] += dp[v][j];
		}
	}
	if (!que[u].size()) return (void) dp[u].push_back(siz[u] - 1);;
	int res = 0;
	for (int i = 1; i <= que[u][0].first; i++)
	{
		res += dp[u][dp[u].size() - i];
	}
	ans[que[u][0].second] = res;
	for (int i = 1; i < que[u].size(); i++)
	{
		for (int j = que[u][i - 1].first + 1; j <= que[u][i].first; j++)
		{
			res += dp[u][dp[u].size() - j];
		}
		ans[que[u][i].second] = res;
	}
	dp[u].push_back(siz[u] - 1);
}

signed main()
{
	memset(head,-1,sizeof(head));
	scanf("%lld%lld",&n,&q);
	for (int i = 1; i < n; i++)
	{
		int u,v;
		scanf("%lld%lld",&u,&v);
		add_edge(u,v);
		add_edge(v,u);
	}
	dfs1(1,0);
	for (int i = 1; i <= q; i++)
	{
		scanf("%lld%lld",&a[i].p,&a[i].k);
		a[i].id = i;
	}
	sort(a + 1,a + q + 1,cmp1);
	for (int i = 1; i <= q; i++)
	{
		que[a[i].p].push_back(make_pair(min(a[i].k,d[a[i].p] - dep[a[i].p]),a[i].id));
	}
	dfs2(1,1);
	sort(a + 1,a + q + 1,cmp2);
	for (int i = 1; i <= q; i++)
	{
		printf("%lld\n",ans[i] + min(dep[a[i].p] - 1,a[i].k) * (siz[a[i].p] - 1));
	}
	return 0;
}