#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long power(long long num, long long r) {
  if (r == 0) return 1;
  if (r == 1) return num % mod;
  long long ans = power(num, r / 2) % mod;
  if (r % 2 == 0) {
    return (ans * ans) % mod;
  }
  return (((ans * ans) % mod) * num) % mod;
}
long long modinv(long long num) { return power(num, mod - 2); }
int findset(int arr[], int x) {
  if (x != arr[x]) {
    arr[x] = findset(arr, arr[x]);
  }
  return arr[x];
}
void uni(int arr[], int ranks[], int a, int b) {
  int ra = findset(arr, a);
  int rb = findset(arr, b);
  if (ra != rb) {
    if (ranks[ra] < ranks[rb]) swap(ra, rb);
    arr[rb] = ra;
    if (ranks[ra] == ranks[rb]) {
      ranks[ra]++;
    }
  }
}
vector<int> vis;
vector<int> d;
void dfs(int node, vector<int> e[]) {
  vis[node] = 1;
  for (int i = 0; i < e[node].size(); i++) {
    int x = e[node][i];
    if (!vis[x]) {
      d[x] = d[node] + 1;
      dfs(x, e);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  if (n == 1) {
    cout << "1 1";
    return 0;
  }
  if (n == 2) {
    cout << "1 1";
    return 0;
  }
  vector<int> e[n + 1];
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  int mx = n - 1;
  int mn = 0;
  for (int i = 1; i <= n; i++) {
    int cnt = 0;
    if (e[i].size() > 1) {
      for (int j = 0; j < e[i].size(); j++) {
        if (e[e[i][j]].size() == 1) cnt++;
      }
    }
    if (cnt > 1) mx -= (cnt - 1);
  }
  int node;
  for (int i = 0; i <= n; i++) {
    vis.push_back(0);
    d.push_back(0);
  }
  for (int i = 1; i <= n; i++) {
    if (e[i].size() > 1) node = i;
  }
  dfs(node, e);
  int a = 0, b = 0;
  for (int i = 1; i <= n; i++) {
    if (e[i].size() == 1) {
      if (d[i] % 2 == 0)
        a++;
      else
        b++;
    }
  }
  if (a == 0 || b == 0)
    mn = 1;
  else
    mn = 3;
  cout << mn << " " << mx;
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to 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 mod = 1e9 + 7;
long long power(long long num, long long r) {
  if (r == 0) return 1;
  if (r == 1) return num % mod;
  long long ans = power(num, r / 2) % mod;
  if (r % 2 == 0) {
    return (ans * ans) % mod;
  }
  return (((ans * ans) % mod) * num) % mod;
}
long long modinv(long long num) { return power(num, mod - 2); }
int findset(int arr[], int x) {
  if (x != arr[x]) {
    arr[x] = findset(arr, arr[x]);
  }
  return arr[x];
}
void uni(int arr[], int ranks[], int a, int b) {
  int ra = findset(arr, a);
  int rb = findset(arr, b);
  if (ra != rb) {
    if (ranks[ra] < ranks[rb]) swap(ra, rb);
    arr[rb] = ra;
    if (ranks[ra] == ranks[rb]) {
      ranks[ra]++;
    }
  }
}
vector<int> vis;
vector<int> d;
void dfs(int node, vector<int> e[]) {
  vis[node] = 1;
  for (int i = 0; i < e[node].size(); i++) {
    int x = e[node][i];
    if (!vis[x]) {
      d[x] = d[node] + 1;
      dfs(x, e);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  if (n == 1) {
    cout << "1 1";
    return 0;
  }
  if (n == 2) {
    cout << "1 1";
    return 0;
  }
  vector<int> e[n + 1];
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  int mx = n - 1;
  int mn = 0;
  for (int i = 1; i <= n; i++) {
    int cnt = 0;
    if (e[i].size() > 1) {
      for (int j = 0; j < e[i].size(); j++) {
        if (e[e[i][j]].size() == 1) cnt++;
      }
    }
    if (cnt > 1) mx -= (cnt - 1);
  }
  int node;
  for (int i = 0; i <= n; i++) {
    vis.push_back(0);
    d.push_back(0);
  }
  for (int i = 1; i <= n; i++) {
    if (e[i].size() > 1) node = i;
  }
  dfs(node, e);
  int a = 0, b = 0;
  for (int i = 1; i <= n; i++) {
    if (e[i].size() == 1) {
      if (d[i] % 2 == 0)
        a++;
      else
        b++;
    }
  }
  if (a == 0 || b == 0)
    mn = 1;
  else
    mn = 3;
  cout << mn << " " << mx;
  return 0;
}

```