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 mxN = 3e5;
const int mxlogN = 19;
const int mod = 1e9 + 7;
int down[mxN];
int cc[mxN];
int up[mxN][mxlogN];
int parent[mxN];
vector<int> children[mxN];
vector<int> adj[mxN];
int precalc(int i, int p = -1) {
  cc[i] = 1;
  for (int j : adj[i]) {
    if (j == p) continue;
    up[j][0] = i;
    down[j] = down[i] + 1;
    cc[i] += precalc(j, i);
  }
  return cc[i];
}
int goup(int u, int k) {
  int j = 0;
  while (k) {
    if (k & 1) u = up[u][j];
    j++;
    k >>= 1;
  }
  return u;
}
int dist(int u, int v) {
  if (down[u] < down[v]) swap(u, v);
  int d = down[u] - down[v];
  u = goup(u, down[u] - down[v]);
  if (u != v) {
    int c = u;
    for (int j = mxlogN - 1; j >= 0; j--) {
      if (up[u][j] != up[v][j]) {
        u = up[u][j];
        v = up[v][j];
      }
    }
    d += 2 * (down[c] - down[u] + 1);
  }
  return d;
}
bool vis[mxN];
int init_centroid(int i, int p = -1) {
  if (vis[i]) {
    cc[i] = 0;
    return 0;
  }
  cc[i] = 1;
  for (int j : adj[i]) {
    if (j == p) continue;
    cc[i] += init_centroid(j, i);
  }
  return cc[i];
}
int is_centroid(int i, int p, int n) {
  for (int j : adj[i]) {
    if (j == p) continue;
    if (cc[j] > n / 2) return j;
  }
  return i;
}
void find_centroid(int i, int p, int n, int x) {
  int centroid = is_centroid(i, p, n);
  if (centroid == i) {
    vis[i] = true;
    parent[i] = x;
    children[x].push_back(i);
    for (int j : adj[i]) {
      if (!vis[j]) {
        init_centroid(j);
        find_centroid(j, -1, cc[j], i);
      }
    }
  } else {
    find_centroid(centroid, i, n, x);
  }
}
int best[mxN];
void update(int i, int n) {
  int j = i;
  while (j != n) {
    best[j] = min(best[j], dist(i, j));
    j = parent[j];
  }
}
int query(int i, int n) {
  int j = i;
  int ans = 1e7;
  while (j != n) {
    ans = min(ans, best[j] + dist(i, j));
    j = parent[j];
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, q;
  cin >> n >> q;
  best[0] = 1e7;
  for (int i = 1; i < n; i++) {
    best[i] = 1e7;
    int a, b;
    cin >> a >> b;
    a--, b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  precalc(0);
  for (int j = 1; j < mxlogN; j++) {
    for (int i = 1; i < n; i++) {
      up[i][j] = up[up[i][j - 1]][j - 1];
    }
  }
  find_centroid(0, -1, n, n);
  update(0, n);
  while (q--) {
    int op, u;
    cin >> op >> u;
    u--;
    if (op == 1) {
      update(u, n);
    } else {
      cout << query(u, n) << "\n";
    }
  }
}
