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 setBit(int n, int pos) { return n = n | (1 << pos); }
int resetBit(int n, int pos) { return n = n & ~(1 << pos); }
bool checkBit(int n, int pos) { return (bool)(n & (1 << pos)); }
vector<int> ed[300010];
bool isCentroid[300010];
int sub[300010];
int cpar[300010];
int clevel[300010];
int dis[20][300010];
void calcSubTree(int s, int par) {
  sub[s] = 1;
  for (auto x : ed[s]) {
    if (x != par && !isCentroid[x]) {
      calcSubTree(x, s);
      sub[s] += sub[x];
    }
  }
}
int nn;
int getCentroid(int s, int par) {
  for (auto x : ed[s]) {
    if (!isCentroid[x] && x != par && sub[x] > (nn / 2))
      return getCentroid(x, s);
  }
  return s;
}
void setDis(int s, int from, int par, int l) {
  dis[from][s] = l;
  for (auto x : ed[s]) {
    if (x != par && !isCentroid[x]) {
      setDis(x, from, s, l + 1);
    }
  }
}
void decompose(int s, int p, int l) {
  calcSubTree(s, p);
  nn = sub[s];
  int c = getCentroid(s, p);
  setDis(c, l, p, 0);
  isCentroid[c] = true;
  cpar[c] = p;
  clevel[c] = l;
  for (auto x : ed[c]) {
    if (!isCentroid[x]) decompose(x, c, l + 1);
  }
}
int ans[300010];
inline void update(int v) {
  int u = v;
  while (u != -1) {
    ans[u] = min(ans[u], dis[clevel[u]][v]);
    u = cpar[u];
  }
}
inline int query(int v) {
  int ret = 2000000000;
  int u = v;
  while (u != -1) {
    ret = min(ret, dis[clevel[u]][v] + ans[u]);
    u = cpar[u];
  }
  return ret;
}
int main() {
  int n, i, u, v, type, q;
  scanf("%d %d", &n, &q);
  for (i = 1; i < n; i++) {
    scanf("%d %d", &u, &v);
    ed[u].push_back(v);
    ed[v].push_back(u);
  }
  decompose(1, -1, 0);
  for (i = 1; i <= n; i++) ans[i] = 2000000000;
  update(1);
  for (i = 1; i <= q; i++) {
    scanf("%d %d", &type, &v);
    if (type == 1)
      update(v);
    else
      printf("%d\n", query(v));
  }
  return 0;
}
