#include <bits/stdc++.h>
using namespace std;
const long long mx = 2e5 + 7;
vector<long long> g[mx];
long long deg[mx];
long long level[mx];
vector<long long> leafs;
bool ok[mx];
void dfs(long long u, long long p) {
  for (long long v : g[u]) {
    if (v != p) {
      level[v] = level[u] + 1;
      dfs(v, u);
    }
  }
}
long long sum;
void dfss(long long u, long long p) {
  long long c = 0;
  for (long long v : g[u]) {
    if (v != p) {
      if (ok[v]) {
        c++;
      }
    }
  }
  if (c > 0) {
    sum += c - 1;
  }
  for (long long v : g[u]) {
    if (v != p) dfss(v, u);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
    deg[u]++, deg[v]++;
  }
  long long root = 1;
  for (long long i = 1; i <= n; i++) {
    if (deg[i] > 1) {
      root = i;
      break;
    }
  }
  level[root] = 1;
  dfs(root, -1);
  for (long long i = 1; i <= n; i++) {
    if (deg[i] == 1) {
      leafs.push_back(i);
      ok[i] = 1;
    }
  }
  long long x = 0, y = 0;
  for (long long u : leafs) {
    if (level[u] % 2)
      x++;
    else
      y++;
  }
  long long minimum = 1;
  if (x > 0 && y > 0) {
    minimum = 3;
  }
  long long maximum = n - 1;
  sum = 0;
  dfss(root, -1);
  maximum -= sum;
  cout << minimum << " " << maximum << endl;
}

### 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 long long mx = 2e5 + 7;
vector<long long> g[mx];
long long deg[mx];
long long level[mx];
vector<long long> leafs;
bool ok[mx];
void dfs(long long u, long long p) {
  for (long long v : g[u]) {
    if (v != p) {
      level[v] = level[u] + 1;
      dfs(v, u);
    }
  }
}
long long sum;
void dfss(long long u, long long p) {
  long long c = 0;
  for (long long v : g[u]) {
    if (v != p) {
      if (ok[v]) {
        c++;
      }
    }
  }
  if (c > 0) {
    sum += c - 1;
  }
  for (long long v : g[u]) {
    if (v != p) dfss(v, u);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
    deg[u]++, deg[v]++;
  }
  long long root = 1;
  for (long long i = 1; i <= n; i++) {
    if (deg[i] > 1) {
      root = i;
      break;
    }
  }
  level[root] = 1;
  dfs(root, -1);
  for (long long i = 1; i <= n; i++) {
    if (deg[i] == 1) {
      leafs.push_back(i);
      ok[i] = 1;
    }
  }
  long long x = 0, y = 0;
  for (long long u : leafs) {
    if (level[u] % 2)
      x++;
    else
      y++;
  }
  long long minimum = 1;
  if (x > 0 && y > 0) {
    minimum = 3;
  }
  long long maximum = n - 1;
  sum = 0;
  dfss(root, -1);
  maximum -= sum;
  cout << minimum << " " << maximum << endl;
}

```