#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10, M = 2e5 + 10;
int n;
int h[N], e[M], ne[M], idx;
vector<int> ru[N];
vector<int> chu[N];
map<int, int> ans;
int dist[N];
bool st[N];
int q[N];
void add(int a, int b) {
  e[idx] = b;
  ne[idx] = h[a];
  h[a] = idx++;
}
void spfa(int s) {
  memset(dist, 0x3f, sizeof dist);
  int hh = 0, tt = 1;
  dist[s] = 0;
  q[0] = s;
  while (hh != tt) {
    int t = q[hh++];
    if (hh == N) hh = 0;
    st[t] = false;
    for (int i = h[t]; ~i; i = ne[i]) {
      int j = e[i];
      if (dist[j] > dist[t] + 1) {
        dist[j] = dist[t] + 1;
        if (!st[j]) {
          st[j] = true;
          q[tt++] = j;
          if (tt == N) tt = 0;
        }
      }
    }
  }
}
int main() {
  cin >> n;
  memset(h, -1, sizeof h);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    ru[a].push_back(b);
    chu[a].push_back(b);
    ru[b].push_back(a);
    chu[b].push_back(a);
    add(a, b);
    add(b, a);
  }
  int m = n - 1;
  int s = 0;
  for (int i = 1; i <= n; i++) {
    if (ru[i].size() == 1 && chu[i].size() == 1) {
      ans[chu[i][0]]++;
      if (!s) s = chu[i][0];
    }
  }
  spfa(s);
  bool ok = true;
  for (auto item : ans) {
    m -= (item.second - 1);
    if (dist[item.first] % 2) ok = false;
  }
  if (ok)
    cout << 1 << ' ' << m << '\n';
  else
    cout << 3 << ' ' << m << '\n';
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int N = 1e5 + 10, M = 2e5 + 10;
int n;
int h[N], e[M], ne[M], idx;
vector<int> ru[N];
vector<int> chu[N];
map<int, int> ans;
int dist[N];
bool st[N];
int q[N];
void add(int a, int b) {
  e[idx] = b;
  ne[idx] = h[a];
  h[a] = idx++;
}
void spfa(int s) {
  memset(dist, 0x3f, sizeof dist);
  int hh = 0, tt = 1;
  dist[s] = 0;
  q[0] = s;
  while (hh != tt) {
    int t = q[hh++];
    if (hh == N) hh = 0;
    st[t] = false;
    for (int i = h[t]; ~i; i = ne[i]) {
      int j = e[i];
      if (dist[j] > dist[t] + 1) {
        dist[j] = dist[t] + 1;
        if (!st[j]) {
          st[j] = true;
          q[tt++] = j;
          if (tt == N) tt = 0;
        }
      }
    }
  }
}
int main() {
  cin >> n;
  memset(h, -1, sizeof h);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    ru[a].push_back(b);
    chu[a].push_back(b);
    ru[b].push_back(a);
    chu[b].push_back(a);
    add(a, b);
    add(b, a);
  }
  int m = n - 1;
  int s = 0;
  for (int i = 1; i <= n; i++) {
    if (ru[i].size() == 1 && chu[i].size() == 1) {
      ans[chu[i][0]]++;
      if (!s) s = chu[i][0];
    }
  }
  spfa(s);
  bool ok = true;
  for (auto item : ans) {
    m -= (item.second - 1);
    if (dist[item.first] % 2) ok = false;
  }
  if (ok)
    cout << 1 << ' ' << m << '\n';
  else
    cout << 3 << ' ' << m << '\n';
  return 0;
}

```