#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10, inf = 0x3f3f3f3f;
int n, Root;
int deg[maxn], dep[maxn];
vector<int> G[maxn];
vector<int> tag;
int sum = 0;
bool leaf[maxn];
inline void add(int x, int y) {
  G[x].push_back(y);
  G[y].push_back(x);
  ++deg[x], ++deg[y];
}
void dfs(int node, int fa) {
  dep[node] = dep[fa] + 1;
  int cnt = 0;
  for (int i = (0); i <= (int)(G[node].size() - 1); ++i) {
    int u = G[node][i];
    if (u == fa) continue;
    dfs(u, node);
    if (leaf[u]) ++cnt;
  }
  if (cnt > 0) sum += cnt - 1;
}
int main() {
  int x, y;
  scanf("%d", &n);
  for (int i = (1); i <= (int)(n - 1); ++i) scanf("%d%d", &x, &y), add(x, y);
  for (int i = (1); i <= (int)(n); ++i) {
    if (deg[i] == 1)
      tag.push_back(i), leaf[i] = 1;
    else if (!Root)
      Root = i;
  }
  dep[0] = -1;
  dfs(Root, 0);
  int odd = 0, even = 0;
  for (int i = (0); i <= (int)(tag.size() - 1); ++i) {
    int u = tag[i];
    if (dep[u] & 1)
      ++odd;
    else
      ++even;
  }
  int Min;
  if (odd > 0 && even > 0)
    Min = 3;
  else
    Min = 1;
  printf("%d %d\n", Min, n - 1 - sum);
  return 0;
}

### 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;
const int maxn = 1e5 + 10, inf = 0x3f3f3f3f;
int n, Root;
int deg[maxn], dep[maxn];
vector<int> G[maxn];
vector<int> tag;
int sum = 0;
bool leaf[maxn];
inline void add(int x, int y) {
  G[x].push_back(y);
  G[y].push_back(x);
  ++deg[x], ++deg[y];
}
void dfs(int node, int fa) {
  dep[node] = dep[fa] + 1;
  int cnt = 0;
  for (int i = (0); i <= (int)(G[node].size() - 1); ++i) {
    int u = G[node][i];
    if (u == fa) continue;
    dfs(u, node);
    if (leaf[u]) ++cnt;
  }
  if (cnt > 0) sum += cnt - 1;
}
int main() {
  int x, y;
  scanf("%d", &n);
  for (int i = (1); i <= (int)(n - 1); ++i) scanf("%d%d", &x, &y), add(x, y);
  for (int i = (1); i <= (int)(n); ++i) {
    if (deg[i] == 1)
      tag.push_back(i), leaf[i] = 1;
    else if (!Root)
      Root = i;
  }
  dep[0] = -1;
  dfs(Root, 0);
  int odd = 0, even = 0;
  for (int i = (0); i <= (int)(tag.size() - 1); ++i) {
    int u = tag[i];
    if (dep[u] & 1)
      ++odd;
    else
      ++even;
  }
  int Min;
  if (odd > 0 && even > 0)
    Min = 3;
  else
    Min = 1;
  printf("%d %d\n", Min, n - 1 - sum);
  return 0;
}

```