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 N = 1e5 + 10;
int n, q, num;
int sz[N], p[N], mark[N], big[N], root[22][N], h[N], best[N];
bool isCent[N];
vector<int> adj[N];
void load() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
}
void dfs(int u) {
  mark[u] = num;
  sz[u] = 1;
  big[u] = 0;
  for (int v : adj[u])
    if (mark[v] != num && !isCent[v]) {
      dfs(v);
      if (sz[v] > sz[big[u]]) {
        big[u] = v;
      }
      sz[u] += sz[v];
    }
}
int findCentroid(int u) {
  num++;
  dfs(u);
  int siz = sz[u] >> 1;
  while (sz[big[u]] > siz) {
    u = big[u];
  }
  isCent[u] = 1;
  for (int v : adj[u])
    if (!isCent[v]) {
      int x = findCentroid(v);
      p[x] = u;
    }
  return u;
}
void dfs(int u, int p) {
  for (int v : adj[u])
    if (v != p) {
      root[0][v] = u;
      for (int i = 1; i < 22; ++i) root[i][v] = root[i - 1][root[i - 1][v]];
      h[v] = h[u] + 1;
      dfs(v, u);
    }
}
int lca(int u, int v) {
  for (int k = 21; k >= 0; --k)
    if (h[root[k][u]] >= h[v]) {
      u = root[k][u];
    }
  for (int k = 21; k >= 0; --k)
    if (h[root[k][v]] >= h[u]) {
      v = root[k][v];
    }
  if (u == v) return u;
  for (int k = 21; k >= 0; --k)
    if (root[k][u] != root[k][v]) {
      u = root[k][u];
      v = root[k][v];
    }
  return root[0][u];
}
int dist(int u, int v) { return h[u] + h[v] - 2 * h[lca(u, v)]; }
void update(int u) {
  int id = u;
  while (u) {
    best[u] = min(best[u], dist(u, id));
    u = p[u];
  }
}
void query(int u) {
  int ans = 1e9 + 7, id = u;
  while (u) {
    ans = min(ans, best[u] + dist(u, id));
    u = p[u];
  }
  printf("%d\n", ans);
}
void process() {
  int r = findCentroid(1);
  dfs(1, 1);
  for (int i = 1; i <= n; ++i) best[i] = 1e9 + 7;
  int x = 1;
  while (x) {
    best[x] = dist(x, 1);
    x = p[x];
  }
  while (q--) {
    int t, u;
    scanf("%d%d", &t, &u);
    if (t == 1) {
      update(u);
    } else {
      query(u);
    }
  }
}
int main() {
  load();
  process();
  return 0;
}
