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;
int N, Q, tot;
set<int> G[100005];
int dp[20][100005], level[100005], pa[100005], subsize[100005], ans[100005];
void dfs(int cur, int par) {
  dp[0][cur] = par;
  for (__typeof(G[cur].begin()) it = G[cur].begin(); it != G[cur].end(); it++) {
    if (*it == par) continue;
    level[*it] = level[cur] + 1;
    dfs(*it, cur);
  }
}
void dfs1(int cur, int par) {
  tot++;
  subsize[cur] = 1;
  for (__typeof(G[cur].begin()) it = G[cur].begin(); it != G[cur].end(); it++) {
    if (*it == par) continue;
    dfs1(*it, cur);
    subsize[cur] += subsize[*it];
  }
}
int dfs2(int cur, int par) {
  for (__typeof(G[cur].begin()) it = G[cur].begin(); it != G[cur].end(); it++) {
    if (*it == par) continue;
    if (subsize[*it] > tot / 2) {
      return dfs2(*it, cur);
    }
  }
  return cur;
}
int lca(int a, int b) {
  if (level[a] > level[b]) swap(a, b);
  int d = level[b] - level[a];
  for (int i = 0; i < 20; i++)
    if (d & (1 << i)) b = dp[i][b];
  if (a == b) return a;
  for (int i = 19; i >= 0; i--)
    if (dp[i][a] != dp[i][b]) a = dp[i][a], b = dp[i][b];
  return dp[0][a];
}
int dist(int u, int v) { return level[u] + level[v] - 2 * level[lca(u, v)]; }
void decompose(int cur, int par) {
  tot = 0;
  dfs1(cur, par);
  int centroid = dfs2(cur, -1);
  pa[centroid] = par;
  for (__typeof(G[centroid].begin()) it = G[centroid].begin();
       it != G[centroid].end(); it++) {
    G[*it].erase(centroid);
    decompose(*it, centroid);
  }
  G[centroid].clear();
}
void update(int u) {
  int x = u;
  while (x != -1) {
    ans[x] = min(ans[x], dist(x, u));
    x = pa[x];
  }
}
int query(int u) {
  int x = u;
  int ret = 1000000000;
  while (x != -1) {
    ret = min(ret, dist(u, x) + ans[x]);
    x = pa[x];
  }
  return ret;
}
int main() {
  int t;
  scanf("%d %d", &N, &Q);
  for (int i = 0; i < N - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--, v--;
    G[u].insert(v);
    G[v].insert(u);
  }
  memset(dp, -1, sizeof dp);
  memset(ans, 0x3f, sizeof ans);
  level[0] = 0;
  dfs(0, -1);
  for (int i = 1; i < 20; i++) {
    for (int j = 0; j < N; j++) {
      if (dp[i - 1][j] != -1) dp[i][j] = dp[i - 1][dp[i - 1][j]];
    }
  }
  decompose(0, -1);
  update(0);
  while (Q--) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--, v--;
    if (u) {
      printf("%d\n", query(v));
    } else {
      update(v);
    }
  }
  return 0;
}
