#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const long long INF = 1e18;
const double eps = 1e-10;
mt19937 mrand(chrono::high_resolution_clock::now().time_since_epoch().count());
template <class T>
inline void upmin(T &x, T y) {
  x = x > y ? y : x;
}
template <class T>
inline void upmax(T &x, T y) {
  x = x < y ? y : x;
}
const int maxn = 1e5 + 233;
vector<int> G[maxn];
int n;
int col[maxn];
int fa[maxn];
int cnt[maxn];
void dfs(int u, int f) {
  for (auto v : G[u]) {
    if (v == f) continue;
    col[v] = 3 - col[u];
    fa[v] = u;
    dfs(v, u);
  }
}
int deg[maxn];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(10);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    G[u].emplace_back(v);
    G[v].emplace_back(u);
    deg[u]++;
    deg[v]++;
  }
  for (int i = 1; i <= n; i++) {
    if (deg[i] > 1) {
      col[i] = 1;
      dfs(i, 0);
      break;
    }
  }
  int f1 = 0, f2 = 0;
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 1) {
      if (col[i] == 1)
        f1 = 1;
      else
        f2 = 1;
      cnt[fa[i]]++;
    }
  }
  int mn, mx;
  if (f1 && f2)
    mn = 3;
  else
    mn = 1;
  mx = n - 1;
  for (int i = 1; i <= n; i++) {
    if (cnt[i]) mx -= cnt[i] - 1;
  }
  cout << mn << ' ' << mx << '\n';
  return 0;
}

### Prompt

Your task is to create 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 inf = 1e9;
const long long INF = 1e18;
const double eps = 1e-10;
mt19937 mrand(chrono::high_resolution_clock::now().time_since_epoch().count());
template <class T>
inline void upmin(T &x, T y) {
  x = x > y ? y : x;
}
template <class T>
inline void upmax(T &x, T y) {
  x = x < y ? y : x;
}
const int maxn = 1e5 + 233;
vector<int> G[maxn];
int n;
int col[maxn];
int fa[maxn];
int cnt[maxn];
void dfs(int u, int f) {
  for (auto v : G[u]) {
    if (v == f) continue;
    col[v] = 3 - col[u];
    fa[v] = u;
    dfs(v, u);
  }
}
int deg[maxn];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(10);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    G[u].emplace_back(v);
    G[v].emplace_back(u);
    deg[u]++;
    deg[v]++;
  }
  for (int i = 1; i <= n; i++) {
    if (deg[i] > 1) {
      col[i] = 1;
      dfs(i, 0);
      break;
    }
  }
  int f1 = 0, f2 = 0;
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 1) {
      if (col[i] == 1)
        f1 = 1;
      else
        f2 = 1;
      cnt[fa[i]]++;
    }
  }
  int mn, mx;
  if (f1 && f2)
    mn = 3;
  else
    mn = 1;
  mx = n - 1;
  for (int i = 1; i <= n; i++) {
    if (cnt[i]) mx -= cnt[i] - 1;
  }
  cout << mn << ' ' << mx << '\n';
  return 0;
}

```