#include <bits/stdc++.h>
using namespace std;
long long MOD = 1000000000;
long long gcdf(long long a, long long b) {
  while (a != 0 && b != 0) {
    if (a > b) {
      a %= b;
    } else {
      b %= a;
    }
  }
  return max(a, b);
}
long long power(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
long long mod_div(long long a, long long b) { return power(a, b - 2, b) % MOD; }
long long solve(long long a, long long b) {
  if (a == 1) return b;
  return (a / b) + solve(b, a % b);
}
vector<vector<int>> v(100001);
bool dfs(int n, int h, int p) {
  int fl = 0, i;
  bool ans = false;
  for (i = 0; i < v[n].size(); i++) {
    if (v[n][i] != p) {
      fl = 1;
      ans = ans | dfs(v[n][i], h + 1, n);
    }
  }
  if (fl == 0) {
    if (h % 2 == 1) ans = true;
  }
  return ans;
}
int main() {
  long long test = 1, i, j, k = 1;
  while (test--) {
    int n;
    cin >> n;
    int inc[100001];
    for (i = 1; i < n; i++) {
      int x, y;
      cin >> x >> y;
      inc[x]++;
      inc[y]++;
      v[x].push_back(y);
      v[y].push_back(x);
    }
    int st = 0;
    for (i = 1; i < n + 1; i++)
      if (inc[i] == 1) st = i;
    bool fir = dfs(st, 0, -1);
    if (fir)
      cout << 3 << " ";
    else
      cout << 1 << " ";
    vector<int> lef;
    int par[100001];
    for (i = 1; i < n + 1; i++) {
      if (inc[i] == 1) {
        par[i] = v[i][0];
        lef.push_back(i);
      }
    }
    long long ans = (n - lef.size()) - 1;
    set<int> uni;
    for (i = 0; i < lef.size(); i++) {
      uni.insert(par[lef[i]]);
    }
    cout << ans + uni.size() << 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 MOD = 1000000000;
long long gcdf(long long a, long long b) {
  while (a != 0 && b != 0) {
    if (a > b) {
      a %= b;
    } else {
      b %= a;
    }
  }
  return max(a, b);
}
long long power(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
long long mod_div(long long a, long long b) { return power(a, b - 2, b) % MOD; }
long long solve(long long a, long long b) {
  if (a == 1) return b;
  return (a / b) + solve(b, a % b);
}
vector<vector<int>> v(100001);
bool dfs(int n, int h, int p) {
  int fl = 0, i;
  bool ans = false;
  for (i = 0; i < v[n].size(); i++) {
    if (v[n][i] != p) {
      fl = 1;
      ans = ans | dfs(v[n][i], h + 1, n);
    }
  }
  if (fl == 0) {
    if (h % 2 == 1) ans = true;
  }
  return ans;
}
int main() {
  long long test = 1, i, j, k = 1;
  while (test--) {
    int n;
    cin >> n;
    int inc[100001];
    for (i = 1; i < n; i++) {
      int x, y;
      cin >> x >> y;
      inc[x]++;
      inc[y]++;
      v[x].push_back(y);
      v[y].push_back(x);
    }
    int st = 0;
    for (i = 1; i < n + 1; i++)
      if (inc[i] == 1) st = i;
    bool fir = dfs(st, 0, -1);
    if (fir)
      cout << 3 << " ";
    else
      cout << 1 << " ";
    vector<int> lef;
    int par[100001];
    for (i = 1; i < n + 1; i++) {
      if (inc[i] == 1) {
        par[i] = v[i][0];
        lef.push_back(i);
      }
    }
    long long ans = (n - lef.size()) - 1;
    set<int> uni;
    for (i = 0; i < lef.size(); i++) {
      uni.insert(par[lef[i]]);
    }
    cout << ans + uni.size() << endl;
  }
}

```