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 = 100005;
int n, q, cnt = 0;
int dis[N], col[N], dp[N][18], depth[N];
vector<int> adj[N], v;
void bfs() {
  memset(dis, -1, sizeof(dis));
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (col[i]) q.push(i), dis[i] = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (auto v : adj[u])
      if (dis[v] == -1) {
        dis[v] = dis[u] + 1;
        q.push(v);
      }
  }
}
void dfs(int u, int parent) {
  dp[u][0] = parent;
  for (auto v : adj[u]) {
    if (v == parent) continue;
    depth[v] = depth[u] + 1;
    dfs(v, u);
  }
}
int lca(int u, int v) {
  if (depth[u] < depth[v]) swap(u, v);
  for (int k = 17; k >= 0; k--) {
    if (depth[u] - (1 << k) >= depth[v]) {
      u = dp[u][k];
    }
  }
  if (u == v) return u;
  for (int k = 17; k >= 0; k--) {
    if (dp[u][k] != dp[v][k]) {
      u = dp[u][k];
      v = dp[v][k];
    }
  }
  return dp[u][0];
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  memset(dp, -1, sizeof(dp));
  dfs(1, -1);
  for (int k = 1; k < 18; k++) {
    for (int u = 1; u <= n; u++) {
      if (dp[u][k - 1] == -1) continue;
      dp[u][k] = dp[dp[u][k - 1]][k - 1];
    }
  }
  col[1] = 1;
  int root = sqrt(q);
  for (int i = 0; i < q; i++) {
    if (i % root == 0) {
      for (int i = 0; i < (int)v.size(); i++) col[v[i]] = 1;
      bfs();
      v.clear();
    }
    int ty, node;
    cin >> ty >> node;
    if (ty == 1)
      v.push_back(node);
    else {
      for (int i = 0; i < (int)v.size(); i++) {
        dis[node] = min(dis[node],
                        depth[v[i]] + depth[node] - 2 * depth[lca(v[i], node)]);
      }
      cout << dis[node] << '\n';
    }
  }
  return 0;
}
