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 INF = 1e9;
const int N = 1e5 + 100;
int n, m, in_centroid[N], sz[N], mn[N];
vector<int> tree[N];
vector<pair<int, int> > centroid[N];
void dfs(int v, int par) {
  sz[v] = 0;
  for (auto i : tree[v]) {
    if (i != par && !in_centroid[v]) {
      dfs(i, v);
      sz[v] += sz[i];
    }
  }
  sz[v]++;
}
void get_dist(int v, int par, int center, int depth) {
  centroid[v].emplace_back(center, depth);
  for (auto i : tree[v]) {
    if (i != par && !in_centroid[i]) {
      get_dist(i, v, center, depth + 1);
    }
  }
}
void build_centroid(int v, int par) {
  dfs(v, -1);
  int p = -1, total = sz[v];
  for (int run = 1; run;) {
    run = 0;
    for (auto i : tree[v]) {
      if (i != p && !in_centroid[i] && sz[i] > total / 2) {
        p = v;
        v = i;
        run = 1;
        break;
      }
    }
  }
  get_dist(v, -1, v, 0);
  in_centroid[v] = 1;
  for (auto i : tree[v]) {
    if (!in_centroid[i]) {
      build_centroid(i, v);
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    tree[a].push_back(b);
    tree[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    mn[i] = INF;
  }
  build_centroid(1, -1);
  for (auto j : centroid[1]) {
    mn[j.first] = j.second;
  }
  for (int i = 0; i < m; i++) {
    int t, v;
    scanf("%d%d", &t, &v);
    if (t == 1) {
      for (auto j : centroid[v]) {
        mn[j.first] = min(mn[j.first], j.second);
      }
    } else {
      int ans = INF;
      for (auto j : centroid[v]) {
        ans = min(ans, j.second + mn[j.first]);
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
