#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long MAXN = 2e5 + 5;
const long long INF = 1e18;
const long double PI = acos((long double)-1);
vector<long long> adjlist[MAXN];
long long depth[MAXN], maxx;
void dfs(long long u, long long par, long long dep) {
  depth[u] = dep;
  for (auto v : adjlist[u]) {
    if (v != par) {
      dfs(v, u, dep + 1);
    }
  }
  return;
}
long long ans2 = 0, output;
set<long long> isleaf;
bool check[MAXN];
void dfs2(long long u, long long par, long long dep) {
  for (auto it : adjlist[u]) {
    if (it == par)
      continue;
    else {
      if (dep == 1) {
        if (isleaf.count(it)) {
          continue;
        } else {
          ans2++;
          dfs2(it, u, dep + 1);
        }
      } else if (dep == 0) {
        ans2++;
        dfs2(it, u, dep + 1);
      } else {
        if (isleaf.count(it))
          check[u] = 1;
        else {
          ans2++;
          dfs2(it, u, dep + 1);
        }
      }
    }
  }
  if (check[u]) ans2++;
  output = ans2;
  return;
}
int main() {
  long long n;
  cin >> n;
  for (int i = (1); i <= (n - 1); i++) {
    long long a, b;
    cin >> a >> b;
    adjlist[a].push_back(b);
    adjlist[b].push_back(a);
  }
  bool odd = 0, even = 0;
  bool depthcheck[MAXN];
  memset(depthcheck, 0, sizeof(depthcheck));
  long long ans1, ans2 = 0;
  long long root;
  for (int i = (1); i <= (n); i++) {
    if (adjlist[i].size() == 1) {
      isleaf.insert(i);
      root = i;
    }
  }
  dfs(root, 1, 0);
  for (auto it : isleaf) {
    if (depth[it] & 1) odd = 1;
  }
  if (odd)
    ans1 = 3;
  else
    ans1 = 1;
  dfs2(root, 1, 0);
  cout << ans1 << " " << output;
}

### Prompt

In CPP, your task is to solve 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 long long MOD = 1e9 + 7;
const long long MAXN = 2e5 + 5;
const long long INF = 1e18;
const long double PI = acos((long double)-1);
vector<long long> adjlist[MAXN];
long long depth[MAXN], maxx;
void dfs(long long u, long long par, long long dep) {
  depth[u] = dep;
  for (auto v : adjlist[u]) {
    if (v != par) {
      dfs(v, u, dep + 1);
    }
  }
  return;
}
long long ans2 = 0, output;
set<long long> isleaf;
bool check[MAXN];
void dfs2(long long u, long long par, long long dep) {
  for (auto it : adjlist[u]) {
    if (it == par)
      continue;
    else {
      if (dep == 1) {
        if (isleaf.count(it)) {
          continue;
        } else {
          ans2++;
          dfs2(it, u, dep + 1);
        }
      } else if (dep == 0) {
        ans2++;
        dfs2(it, u, dep + 1);
      } else {
        if (isleaf.count(it))
          check[u] = 1;
        else {
          ans2++;
          dfs2(it, u, dep + 1);
        }
      }
    }
  }
  if (check[u]) ans2++;
  output = ans2;
  return;
}
int main() {
  long long n;
  cin >> n;
  for (int i = (1); i <= (n - 1); i++) {
    long long a, b;
    cin >> a >> b;
    adjlist[a].push_back(b);
    adjlist[b].push_back(a);
  }
  bool odd = 0, even = 0;
  bool depthcheck[MAXN];
  memset(depthcheck, 0, sizeof(depthcheck));
  long long ans1, ans2 = 0;
  long long root;
  for (int i = (1); i <= (n); i++) {
    if (adjlist[i].size() == 1) {
      isleaf.insert(i);
      root = i;
    }
  }
  dfs(root, 1, 0);
  for (auto it : isleaf) {
    if (depth[it] & 1) odd = 1;
  }
  if (odd)
    ans1 = 3;
  else
    ans1 = 1;
  dfs2(root, 1, 0);
  cout << ans1 << " " << output;
}

```