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:

  1. paint a specified blue node in red; 
  2. 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.

Input

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.

Output

For each second type query print the reply in a single line.

Examples

Input

5 4
1 2
2 3
2 4
4 5
2 1
2 5
1 2
2 5


Output

0
3
2
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 5;
const int MOD = 1e9 + 7;
const int LOGN = 21;
const int buck = 400;
int n, m;
vector<int> G[MAX];
int par[MAX], lev[MAX];
int P[LOGN][MAX];
vector<int> reds;
int dis[MAX];
void bfs() {
  queue<int> Q;
  for (int v : reds) {
    Q.push(v);
    dis[v] = 0;
  }
  reds.clear();
  while (Q.empty() == false) {
    int u = Q.front();
    Q.pop();
    for (int v : G[u]) {
      if (dis[v] > 1 + dis[u]) {
        dis[v] = 1 + dis[u];
        Q.push(v);
      }
    }
  }
}
void dfs(int u, int p, int lv) {
  par[u] = p, lev[u] = lv;
  for (int v : G[u]) {
    if (v != p) {
      dfs(v, u, lv + 1);
    }
  }
}
void makeSparse() {
  for (int i = 1; i <= n; i++) {
    P[0][i] = par[i];
  }
  for (int i = 1; i < LOGN; i++) {
    for (int j = 1; j <= n; j++) {
      if (P[i - 1][j] != -1) {
        P[i][j] = P[i - 1][P[i - 1][j]];
      }
    }
  }
}
int lca(int u, int v) {
  if (u == v) {
    return u;
  }
  if (lev[u] < lev[v]) {
    swap(u, v);
  }
  for (int i = LOGN - 1; i >= 0; i--) {
    if (lev[u] - (1 << i) >= lev[v]) {
      u = P[i][u];
    }
  }
  if (u == v) {
    return u;
  }
  for (int i = LOGN - 1; i >= 0; i--) {
    if (P[i][u] != P[i][v]) {
      u = P[i][u];
      v = P[i][v];
    }
  }
  return par[u];
}
int dist(int u, int v) { return lev[u] + lev[v] - 2 * lev[lca(u, v)]; }
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, -1, 0);
  makeSparse();
  fill(dis, dis + MAX, (int)1e8);
  reds.push_back(1);
  for (int i = 0; i < m; i++) {
    int t, v;
    scanf("%d %d", &t, &v);
    if (i % buck == 0) {
      bfs();
    }
    if (t == 1) {
      reds.push_back(v);
    } else {
      int ans = dis[v];
      for (int u : reds) {
        ans = min(ans, dist(u, v));
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
