#include <bits/stdc++.h>
using namespace std;
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
const long long MxN = 100005;
long long vis[MxN] = {0};
long long dis[MxN] = {0};
vector<int> adj[MxN];
long long mn, mx;
bool bfs(long long u) {
  vis[u] = 1;
  queue<long long> q;
  q.push(u);
  while (!q.empty()) {
    long long v = q.front();
    q.pop();
    bool flag = 0;
    for (auto it : adj[v]) {
      if (vis[it] == 0) {
        flag = 1;
        dis[it] = dis[v] + 1;
        vis[it] = 1;
        q.push(it);
      }
    }
    if (flag == 0 and dis[v] % 2) return 1;
  }
  return 0;
}
void bfs2(long long u) {
  vis[u] = 1;
  queue<long long> q;
  q.push(u);
  while (!q.empty()) {
    long long v = q.front();
    q.pop();
    bool flag = 0;
    for (auto it : adj[v]) {
      if (vis[it] == 0) {
        vis[it] = 1;
        q.push(it);
        if (adj[it].size() == 1 and flag == 1)
          mx--;
        else if (adj[it].size() == 1)
          flag = 1;
      }
    }
  }
}
bool test = 0;
void solve() {
  long long n, i, x, y;
  cin >> n;
  for (long long i = (0); i < (n - 1); i++) {
    cin >> x >> y;
    adj[x].emplace_back(y);
    adj[y].emplace_back(x);
  }
  for (long long i = (1); i < (n + 1); i++) {
    if (adj[i].size() == 1) {
      bool f = bfs(i);
      if (f)
        mn = 3;
      else
        mn = 1;
      break;
    }
  }
  mx = n - 1;
  for (long long i = (1); i < (n + 1); i++) vis[i] = 0;
  for (long long i = (1); i < (n + 1); i++) {
    if (adj[i].size() > 1) {
      bfs2(i);
      break;
    }
  }
  cout << mn << " " << mx << endl;
}
int main() {
  int t = 1;
  if (test) cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:

  * For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0. 



Note that you can put very large positive integers (like 10^{(10^{10})}).

It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.

<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).

<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0. 

What are the minimum and the maximum possible values of f for the given tree? Find and print both.

Input

The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.

The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.

Output

Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.

Examples

Input


6
1 3
2 3
3 4
4 5
5 6


Output


1 4


Input


6
1 3
2 3
3 4
4 5
4 6


Output


3 3


Input


7
1 2
2 7
3 4
4 7
5 6
6 7


Output


1 6

Note

In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum. 

<image>

In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3. 

<image>

In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum. 

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
const long long MxN = 100005;
long long vis[MxN] = {0};
long long dis[MxN] = {0};
vector<int> adj[MxN];
long long mn, mx;
bool bfs(long long u) {
  vis[u] = 1;
  queue<long long> q;
  q.push(u);
  while (!q.empty()) {
    long long v = q.front();
    q.pop();
    bool flag = 0;
    for (auto it : adj[v]) {
      if (vis[it] == 0) {
        flag = 1;
        dis[it] = dis[v] + 1;
        vis[it] = 1;
        q.push(it);
      }
    }
    if (flag == 0 and dis[v] % 2) return 1;
  }
  return 0;
}
void bfs2(long long u) {
  vis[u] = 1;
  queue<long long> q;
  q.push(u);
  while (!q.empty()) {
    long long v = q.front();
    q.pop();
    bool flag = 0;
    for (auto it : adj[v]) {
      if (vis[it] == 0) {
        vis[it] = 1;
        q.push(it);
        if (adj[it].size() == 1 and flag == 1)
          mx--;
        else if (adj[it].size() == 1)
          flag = 1;
      }
    }
  }
}
bool test = 0;
void solve() {
  long long n, i, x, y;
  cin >> n;
  for (long long i = (0); i < (n - 1); i++) {
    cin >> x >> y;
    adj[x].emplace_back(y);
    adj[y].emplace_back(x);
  }
  for (long long i = (1); i < (n + 1); i++) {
    if (adj[i].size() == 1) {
      bool f = bfs(i);
      if (f)
        mn = 3;
      else
        mn = 1;
      break;
    }
  }
  mx = n - 1;
  for (long long i = (1); i < (n + 1); i++) vis[i] = 0;
  for (long long i = (1); i < (n + 1); i++) {
    if (adj[i].size() > 1) {
      bfs2(i);
      break;
    }
  }
  cout << mn << " " << mx << endl;
}
int main() {
  int t = 1;
  if (test) cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```