Xenia the programmer has a tree consisting of n nodes. We will consider the tree nodes indexed from 1 to n. We will also consider the first node to be initially painted red, and the other nodes — to be painted blue.
The distance between two tree nodes v and u is the number of edges in the shortest path between v and u.
Xenia needs to learn how to quickly execute queries of two types:
- paint a specified blue node in red;
- calculate which red node is the closest to the given one and print the shortest distance to the closest red node.
Your task is to write a program which will execute the described queries.
The first line contains two integers n and m (2 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of nodes in the tree and the number of queries. Next n - 1 lines contain the tree edges, the i-th line contains a pair of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — an edge of the tree.
Next m lines contain queries. Each query is specified as a pair of integers ti, vi (1 ≤ ti ≤ 2, 1 ≤ vi ≤ n). If ti = 1, then as a reply to the query we need to paint a blue node vi in red. If ti = 2, then we should reply to the query by printing the shortest distance from some red node to node vi.
It is guaranteed that the given graph is a tree and that all queries are correct.
For each second type query print the reply in a single line.
5 4 1 2 2 3 2 4 4 5 2 1 2 5 1 2 2 5
0 3 2
#define happy #include<bits/stdc++.h> using namespace std; #define ll long long #define rep(i,a,b) for(int i=a;i<=b;i++) #define all(a) (a).begin(),(a).end() #define pll pair<ll,ll> #define vi vector<int> #define pb push_back const int inf=0x3f3f3f3f; ll rd(){ ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } const int N=1e5+10; int dis[N],head[N],nxt[N<<1],v[N<<1],qhead,qtail,q[N]; bool red[N]; void bfs(int s){ qhead=0;qtail=1; q[1]=s; dis[s]=0; while(qhead<qtail){ int i=q[++qhead]; for(int j=head[i];~j;j=nxt[j]){ int u=v[j]; if(dis[u]>dis[i]+1){ dis[u]=dis[i]+1; q[++qtail]=u; } } } } int vc[N]; int rdis[N]; int rbfs(int s,int st){ if(red[s])return 0; qhead=0;qtail=1; q[1]=s; vc[s]=st; rdis[s]=0; while(qhead<qtail){ int i=q[++qhead]; for(int j=head[i];~j;j=nxt[j]){ int u=v[j]; if(vc[u]!=st){ vc[u]=st; rdis[u]=rdis[i]+1; if(red[u]){ return rdis[u]; } q[++qtail]=u; } } } } int main(){ #ifdef happy freopen("in.txt","r",stdin); #endif int top=0; memset(head,-1,sizeof(head)); memset(vc,-1,sizeof(vc)); int n=rd(),m=rd(); rep(i,1,n-1){ int x=rd()-1,y=rd()-1; v[top]=y,nxt[top]=head[x],head[x]=top;++top; v[top]=x,nxt[top]=head[y],head[y]=top;++top; } rep(i,0,n-1)dis[i]=inf; red[0]=1; bfs(0); int cnt=1; rep(i,0,m-1){ int x=rd(),y=rd()-1; if(x==1){ red[y]=1; ++cnt; if(cnt<=300){ bfs(y); } }else{ if(cnt<=300)printf("%d\n",dis[y]); else printf("%d\n",rbfs(y,i)); } } }