#include <bits/stdc++.h>
using namespace std;
const int INF = (int)2e9;
const long long INFL = (long long)9e18;
const int MAXINT = ((~0) ^ (1 << 31));
const long long MAXLL = ((~0) ^ ((long long)1 << 63));
template <class T>
inline T pow2(T a) {
  return a * a;
}
template <class T>
inline bool mineq(T& a, T b) {
  return (a > b) ? (a = b, true) : false;
}
template <class T>
inline bool maxeq(T& a, T b) {
  return (a < b) ? (a = b, true) : false;
}
const int maxn = 2e5;
long long n, v, u, mark[maxn], dp[maxn][2], flg, ans;
vector<long long> g[maxn];
void dfs(long long v) {
  mark[v] = 1;
  long long cnt = 0;
  for (int i = 0; i < g[v].size(); i++) {
    if (!mark[g[v][i]]) {
      if (g[g[v][i]].size() == 1) {
        cnt++;
        dp[v][1]++;
      }
      dfs(g[v][i]);
      dp[v][0] += dp[g[v][i]][1];
      dp[v][1] += dp[g[v][i]][0];
    }
  }
  ans -= max((long long)0, cnt - 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> v >> u;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  ans = n - 1;
  for (int i = 1; i <= n; i++) {
    if (g[i].size() != 1) {
      dfs(i);
      if (dp[i][0] && dp[i][1]) flg = 1;
      break;
    }
  }
  if (flg)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  cout << ans;
}

### 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 = (int)2e9;
const long long INFL = (long long)9e18;
const int MAXINT = ((~0) ^ (1 << 31));
const long long MAXLL = ((~0) ^ ((long long)1 << 63));
template <class T>
inline T pow2(T a) {
  return a * a;
}
template <class T>
inline bool mineq(T& a, T b) {
  return (a > b) ? (a = b, true) : false;
}
template <class T>
inline bool maxeq(T& a, T b) {
  return (a < b) ? (a = b, true) : false;
}
const int maxn = 2e5;
long long n, v, u, mark[maxn], dp[maxn][2], flg, ans;
vector<long long> g[maxn];
void dfs(long long v) {
  mark[v] = 1;
  long long cnt = 0;
  for (int i = 0; i < g[v].size(); i++) {
    if (!mark[g[v][i]]) {
      if (g[g[v][i]].size() == 1) {
        cnt++;
        dp[v][1]++;
      }
      dfs(g[v][i]);
      dp[v][0] += dp[g[v][i]][1];
      dp[v][1] += dp[g[v][i]][0];
    }
  }
  ans -= max((long long)0, cnt - 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> v >> u;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  ans = n - 1;
  for (int i = 1; i <= n; i++) {
    if (g[i].size() != 1) {
      dfs(i);
      if (dp[i][0] && dp[i][1]) flg = 1;
      break;
    }
  }
  if (flg)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  cout << ans;
}

```