#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 111;
int list[maxn];
vector<int> G[maxn];
set<int> ins;
int vis[maxn];
int z;
int head[maxn];
void add(int x, int y) { G[x].push_back(y); }
long long dp[maxn];
int siz[maxn];
int de[maxn];
int dep[maxn];
int dfs(int x, int fa) {
  for (int p : G[x]) {
    if (p == fa) continue;
    dep[p] = dep[x] + 1;
    if (G[p].size() == 1) {
      vis[dep[p] & 1]++;
    } else
      dfs(p, x);
  }
  return 0;
}
int main() {
  int n;
  scanf("%d", &n);
  int be, en;
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &be, &en);
    add(be, en);
    add(en, be);
    de[be]++;
    de[en]++;
  }
  long long ans1 = n - 1;
  int cnt = 0;
  int root = 0;
  for (int i = 1; i <= n; i++) {
    cnt = 0;
    for (int j = 0; j < G[i].size(); j++) {
      if (G[i].size() >= 2) root = i;
      if (de[G[i][j]] == 1) cnt++;
    }
    if (cnt >= 2) {
      ans1 -= cnt - 1;
    }
  }
  dfs(root, 0);
  int ans = 0;
  if (vis[0] == 0 || vis[1] == 0)
    ans = 1;
  else
    ans = 3;
  cout << ans << " " << ans1 << endl;
  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;
const int maxn = 1e5 + 111;
int list[maxn];
vector<int> G[maxn];
set<int> ins;
int vis[maxn];
int z;
int head[maxn];
void add(int x, int y) { G[x].push_back(y); }
long long dp[maxn];
int siz[maxn];
int de[maxn];
int dep[maxn];
int dfs(int x, int fa) {
  for (int p : G[x]) {
    if (p == fa) continue;
    dep[p] = dep[x] + 1;
    if (G[p].size() == 1) {
      vis[dep[p] & 1]++;
    } else
      dfs(p, x);
  }
  return 0;
}
int main() {
  int n;
  scanf("%d", &n);
  int be, en;
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &be, &en);
    add(be, en);
    add(en, be);
    de[be]++;
    de[en]++;
  }
  long long ans1 = n - 1;
  int cnt = 0;
  int root = 0;
  for (int i = 1; i <= n; i++) {
    cnt = 0;
    for (int j = 0; j < G[i].size(); j++) {
      if (G[i].size() >= 2) root = i;
      if (de[G[i][j]] == 1) cnt++;
    }
    if (cnt >= 2) {
      ans1 -= cnt - 1;
    }
  }
  dfs(root, 0);
  int ans = 0;
  if (vis[0] == 0 || vis[1] == 0)
    ans = 1;
  else
    ans = 3;
  cout << ans << " " << ans1 << endl;
  return 0;
}

```