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 + 7;
const int M = 1001;
const int inf = 1e9 + 7;
const int oth = 853454;
const double pi = acos(-1);
vector<int> adj[N];
int n;
int f[N];
int dist[N];
int pa[N];
int c[N];
int cur = 0;
void DFS(int u, int r) {
  f[u] = 1;
  for (int j = (0); j < (adj[u].size()); j++) {
    int v = adj[u][j];
    if (v != r && pa[v] == 0) {
      DFS(v, u);
      f[u] += f[v];
    }
  }
}
int d[N][25];
int curDepth;
void travel(int u, int r) {
  for (int j = (0); j < (adj[u].size()); j++) {
    int v = adj[u][j];
    if (v != r && pa[v] == 0) {
      d[v][curDepth] = d[u][curDepth] + 1;
      travel(v, u);
    }
  }
}
void build_centroid(int u, int r) {
  DFS(u, r);
  int m = f[u];
  int pre = r;
  while (true) {
    int nxt = -1;
    for (int j = (0); j < (adj[u].size()); j++) {
      int v = adj[u][j];
      if (v != pre && pa[v] == 0) {
        if (f[v] > m / 2) {
          nxt = v;
          break;
        }
      }
    }
    if (nxt == -1) break;
    pre = u;
    u = nxt;
  }
  pa[u] = r;
  if (r != -1) dist[u] = dist[r] + 1;
  curDepth = dist[u];
  for (int j = (0); j < (adj[u].size()); j++) {
    int v = adj[u][j];
    if (pa[v] == 0) {
      d[v][curDepth] = 1;
      travel(v, u);
    }
  }
  for (int j = (0); j < (adj[u].size()); j++) {
    int v = adj[u][j];
    if (pa[v] == 0) {
      build_centroid(v, u);
    }
  }
}
int ans[N];
int LCA_centroid(int u, int v) {
  if (dist[u] < dist[v]) swap(u, v);
  while (dist[u] > dist[v]) u = pa[u];
  if (u == v) return u;
  while (u != v) {
    u = pa[u];
    v = pa[v];
  }
  return u;
}
int path(int u, int v) {
  int c = LCA_centroid(u, v);
  return d[u][dist[c]] + d[v][dist[c]];
}
void update(int u) {
  int r = u;
  while (r != -1) {
    ans[r] = min(ans[r], path(u, r));
    r = pa[r];
  }
}
int get(int u) {
  int r = u;
  int res = inf;
  while (r != -1) {
    res = min(res, ans[r] + path(u, r));
    r = pa[r];
  }
  return res;
}
void solve() {
  int q;
  cin >> 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);
  }
  build_centroid(1, -1);
  for (int i = 1; i <= n; i++) ans[i] = inf;
  update(1);
  while (q-- > 0) {
    int op, u;
    scanf("%d%d", &op, &u);
    if (op == 1) {
      update(u);
    } else {
      printf("%d\n", get(u));
    }
  }
}
int main() { solve(); }
