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 MAXN = 1e5 + 5;
int dp[MAXN][25];
int h[MAXN];
int d[MAXN];
vector<int> v1[MAXN];
vector<int> v2;
void dfs(int curr, int par) {
  dp[curr][0] = par;
  for (int x : v1[curr]) {
    if (x != par) {
      h[x] = h[curr] + 1;
      dfs(x, curr);
    }
  }
}
int lca(int x, int y) {
  if (h[x] > h[y]) {
    swap(x, y);
  }
  for (int i = 19; i >= 0; i--) {
    if (h[y] - (1 << i) >= h[x]) {
      y = dp[y][i];
    }
  }
  if (x == y) {
    return x;
  }
  for (int i = 20; i >= 0; i--) {
    if (dp[x][i] != -1 && dp[x][i] != dp[y][i]) {
      x = dp[x][i];
      y = dp[y][i];
    }
  }
  return dp[x][0];
}
void bfs() {
  queue<int> q1;
  for (int x : v2) {
    q1.push(x);
    d[x] = 0;
  }
  v2.clear();
  while (!q1.empty()) {
    int curr = q1.front();
    q1.pop();
    for (int x : v1[curr]) {
      if (d[x] > d[curr] + 1) {
        d[x] = d[curr] + 1;
        q1.push(x);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    v1[x].push_back(y);
    v1[y].push_back(x);
  }
  for (int i = 0; i <= 19; i++) {
    for (int j = 0; j <= n; j++) {
      dp[j][i] = -1;
      d[j] = 1e9;
    }
  }
  dfs(1, -1);
  for (int i = 1; i <= 19; i++) {
    for (int j = 1; j <= n; j++) {
      if (dp[j][i - 1] == -1) {
        dp[j][i] = -1;
      } else {
        dp[j][i] = dp[dp[j][i - 1]][i - 1];
      }
    }
  }
  d[1] = 0;
  v2.push_back(1);
  bfs();
  int block_size = sqrt(m);
  for (int i = 0; i < m; i += block_size) {
    for (int j = i; j < i + block_size && j < m; j++) {
      int x, y;
      cin >> x >> y;
      if (x == 1) {
        v2.push_back(y);
      } else {
        int ans = d[y];
        for (int z : v2) {
          ans = min(ans, h[y] + h[z] - 2 * h[lca(y, z)]);
        }
        cout << ans << endl;
      }
    }
    bfs();
  }
}
