#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
const ll INF = 1e18;
const ll MOD = 1e9 + 7;
vector<vector<int>> G;
vector<int> colors;
bool flag = false;
int c = 1;
vector<int> tmp;
void dfs(int v, int p, int d) {
  if (d >= 2) return;
  for (auto nv : G[v]) {
    if (nv == p) continue;
    if (G[nv].size() == 1) {
      flag = true;
      tmp.push_back(nv);
    }
    dfs(nv, v, d + 1);
  }
}
int main() {
  int n;
  cin >> n;
  G.assign(n + 1, vector<int>());
  colors.resize(n + 1);
  for (int i = 0; i < (int)(n - 1); i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    if (colors[i] == 0 && G[i].size() == 1) {
      flag = false;
      tmp.clear();
      dfs(i, -1, 0);
      if (tmp.size() != 0) colors[i] = c;
      for (int j = 0; j < (int)(tmp.size()); j++) colors[tmp[j]] = c;
      if (flag) c++;
    }
  }
  int maxv = n - 1;
  map<int, int> m;
  for (int i = 1; i <= n; i++) {
    if (colors[i] != 0) m[colors[i]]++;
  }
  for (auto p : m) {
    maxv -= p.second;
    maxv += 1;
  }
  vector<int> dist(n + 1);
  function<void(int, int, int)> rec = [&](int v, int p, int d) {
    dist[v] = d;
    for (auto nv : G[v]) {
      if (nv == p) continue;
      rec(nv, v, d + 1);
    }
  };
  int root = 0;
  for (int i = 1; i <= n; i++) {
    if (G[i].size() == 1) root = i;
  }
  rec(root, -1, 0);
  bool parity = false;
  for (int i = 1; i <= n; i++) {
    if (dist[i] % 2 == 1 && G[i].size() == 1) parity = true;
  }
  int minv = 1;
  if (parity) minv = 3;
  cout << minv << " " << maxv << endl;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
using ll = long long;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
const ll INF = 1e18;
const ll MOD = 1e9 + 7;
vector<vector<int>> G;
vector<int> colors;
bool flag = false;
int c = 1;
vector<int> tmp;
void dfs(int v, int p, int d) {
  if (d >= 2) return;
  for (auto nv : G[v]) {
    if (nv == p) continue;
    if (G[nv].size() == 1) {
      flag = true;
      tmp.push_back(nv);
    }
    dfs(nv, v, d + 1);
  }
}
int main() {
  int n;
  cin >> n;
  G.assign(n + 1, vector<int>());
  colors.resize(n + 1);
  for (int i = 0; i < (int)(n - 1); i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    if (colors[i] == 0 && G[i].size() == 1) {
      flag = false;
      tmp.clear();
      dfs(i, -1, 0);
      if (tmp.size() != 0) colors[i] = c;
      for (int j = 0; j < (int)(tmp.size()); j++) colors[tmp[j]] = c;
      if (flag) c++;
    }
  }
  int maxv = n - 1;
  map<int, int> m;
  for (int i = 1; i <= n; i++) {
    if (colors[i] != 0) m[colors[i]]++;
  }
  for (auto p : m) {
    maxv -= p.second;
    maxv += 1;
  }
  vector<int> dist(n + 1);
  function<void(int, int, int)> rec = [&](int v, int p, int d) {
    dist[v] = d;
    for (auto nv : G[v]) {
      if (nv == p) continue;
      rec(nv, v, d + 1);
    }
  };
  int root = 0;
  for (int i = 1; i <= n; i++) {
    if (G[i].size() == 1) root = i;
  }
  rec(root, -1, 0);
  bool parity = false;
  for (int i = 1; i <= n; i++) {
    if (dist[i] % 2 == 1 && G[i].size() == 1) parity = true;
  }
  int minv = 1;
  if (parity) minv = 3;
  cout << minv << " " << maxv << endl;
}

```