#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int minn(long long int a, long long int b) {
  if (a > b) return b;
  return a;
}
vector<int> v[100005];
int min_dfs(int s, int par = -1, int prev = 0) {
  int tmp, temp;
  if (v[s].size() == 1 && v[s][0] == par) {
    return prev;
  }
  if (par != -1 && v[s].size() == 2) {
    if (v[s][0] == par)
      tmp = v[s][1];
    else
      tmp = v[s][0];
    return min_dfs(tmp, s, prev + 1);
  }
  set<bool> st;
  st.clear();
  for (int vv : v[s]) {
    if (vv == par) continue;
    temp = min_dfs(vv, s, +1);
    if (temp == -1) return -1;
    st.insert(temp & 1);
  }
  if (st.size() == 2) {
    return -1;
  } else {
    tmp = *(st.begin());
    tmp ^= (prev & 1);
    return tmp;
  }
}
int max_dfs(int s, int par = -1, int prev = 0) {
  int tmp, temp, ans = 0;
  if (v[s].size() == 1 && v[s][0] == par) {
    return prev;
  }
  if (par != -1 && v[s].size() == 2) {
    if (v[s][0] == par)
      tmp = v[s][1];
    else
      tmp = v[s][0];
    return max_dfs(tmp, s, prev + 1);
  }
  set<bool> st;
  st.clear();
  bool ongot = 0;
  for (int vv : v[s]) {
    if (vv == par) continue;
    temp = max_dfs(vv, s, +1);
    if (temp == 1)
      ongot = 1;
    else
      ans += temp;
  }
  if (ongot) ans++;
  return ans + prev;
}
int main() {
  int n;
  while (cin >> n) {
    int i, vv, uu;
    for (i = 0; i < 100005; i++) v[i].clear();
    for (i = 1; i < n; i++) {
      cin >> uu >> vv;
      v[uu].push_back(vv);
      v[vv].push_back(uu);
    }
    for (i = 1; i < n; i++) {
      if (v[i].size() > 1) break;
    }
    int min_ans = min_dfs(i);
    if (min_ans == -1)
      min_ans = 3;
    else
      min_ans = 1;
    int max_ans = max_dfs(i);
    cout << min_ans << ' ' << max_ans << endl;
  }
}

### 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;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int minn(long long int a, long long int b) {
  if (a > b) return b;
  return a;
}
vector<int> v[100005];
int min_dfs(int s, int par = -1, int prev = 0) {
  int tmp, temp;
  if (v[s].size() == 1 && v[s][0] == par) {
    return prev;
  }
  if (par != -1 && v[s].size() == 2) {
    if (v[s][0] == par)
      tmp = v[s][1];
    else
      tmp = v[s][0];
    return min_dfs(tmp, s, prev + 1);
  }
  set<bool> st;
  st.clear();
  for (int vv : v[s]) {
    if (vv == par) continue;
    temp = min_dfs(vv, s, +1);
    if (temp == -1) return -1;
    st.insert(temp & 1);
  }
  if (st.size() == 2) {
    return -1;
  } else {
    tmp = *(st.begin());
    tmp ^= (prev & 1);
    return tmp;
  }
}
int max_dfs(int s, int par = -1, int prev = 0) {
  int tmp, temp, ans = 0;
  if (v[s].size() == 1 && v[s][0] == par) {
    return prev;
  }
  if (par != -1 && v[s].size() == 2) {
    if (v[s][0] == par)
      tmp = v[s][1];
    else
      tmp = v[s][0];
    return max_dfs(tmp, s, prev + 1);
  }
  set<bool> st;
  st.clear();
  bool ongot = 0;
  for (int vv : v[s]) {
    if (vv == par) continue;
    temp = max_dfs(vv, s, +1);
    if (temp == 1)
      ongot = 1;
    else
      ans += temp;
  }
  if (ongot) ans++;
  return ans + prev;
}
int main() {
  int n;
  while (cin >> n) {
    int i, vv, uu;
    for (i = 0; i < 100005; i++) v[i].clear();
    for (i = 1; i < n; i++) {
      cin >> uu >> vv;
      v[uu].push_back(vv);
      v[vv].push_back(uu);
    }
    for (i = 1; i < n; i++) {
      if (v[i].size() > 1) break;
    }
    int min_ans = min_dfs(i);
    if (min_ans == -1)
      min_ans = 3;
    else
      min_ans = 1;
    int max_ans = max_dfs(i);
    cout << min_ans << ' ' << max_ans << endl;
  }
}

```