#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
long long a[N];
vector<long long> node[N];
long long dp[N][20], dep[N], vis[N];
void dfs(long long u, long long par) {
  for (int i = 1; i < 20; i++) {
    dp[u][i] = dp[dp[u][i - 1]][i - 1];
  }
  for (auto j : node[u]) {
    if (j != par) {
      dep[j] = dep[u] + 1;
      dp[j][0] = u;
      dfs(j, u);
    }
  }
}
long long lca(long long x, long long y) {
  if (dep[x] > dep[y]) swap(x, y);
  long long d = dep[y] - dep[x];
  for (int i = 0; i < 20; i++) {
    if (d >> i & 1) {
      y = dp[y][i];
    }
  }
  if (x == y) return x;
  for (int i = 20 - 1; i >= 0; i--) {
    if (dp[x][i] != dp[y][i]) {
      x = dp[x][i];
      y = dp[y][i];
    }
  }
  return dp[x][0];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    node[x].push_back(y);
    node[y].push_back(x);
  }
  vector<long long> one;
  for (int i = 1; i <= n; i++) {
    if (node[i].size() == 1) {
      one.push_back(i);
      vis[i] = 1;
    }
  }
  dfs(one[0], one[0]);
  long long ans = 1;
  for (int i = 1; i < one.size(); i++) {
    long long x = dep[one[0]] + dep[one[i]] - dep[lca(one[0], one[i])];
    if (x % 2 == 1) {
      ans = 3;
    }
  }
  long long ans2 = n - 1;
  for (int i = 1; i <= n; i++) {
    long long cnt = 0;
    for (auto j : node[i]) {
      if (vis[j] == 1) {
        cnt++;
      }
    }
    if (cnt >= 2) ans2 = ans2 - cnt + 1;
  }
  cout << ans << " " << ans2;
}

### Prompt

Your challenge is to write 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 N = 2e5 + 5;
long long a[N];
vector<long long> node[N];
long long dp[N][20], dep[N], vis[N];
void dfs(long long u, long long par) {
  for (int i = 1; i < 20; i++) {
    dp[u][i] = dp[dp[u][i - 1]][i - 1];
  }
  for (auto j : node[u]) {
    if (j != par) {
      dep[j] = dep[u] + 1;
      dp[j][0] = u;
      dfs(j, u);
    }
  }
}
long long lca(long long x, long long y) {
  if (dep[x] > dep[y]) swap(x, y);
  long long d = dep[y] - dep[x];
  for (int i = 0; i < 20; i++) {
    if (d >> i & 1) {
      y = dp[y][i];
    }
  }
  if (x == y) return x;
  for (int i = 20 - 1; i >= 0; i--) {
    if (dp[x][i] != dp[y][i]) {
      x = dp[x][i];
      y = dp[y][i];
    }
  }
  return dp[x][0];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    node[x].push_back(y);
    node[y].push_back(x);
  }
  vector<long long> one;
  for (int i = 1; i <= n; i++) {
    if (node[i].size() == 1) {
      one.push_back(i);
      vis[i] = 1;
    }
  }
  dfs(one[0], one[0]);
  long long ans = 1;
  for (int i = 1; i < one.size(); i++) {
    long long x = dep[one[0]] + dep[one[i]] - dep[lca(one[0], one[i])];
    if (x % 2 == 1) {
      ans = 3;
    }
  }
  long long ans2 = n - 1;
  for (int i = 1; i <= n; i++) {
    long long cnt = 0;
    for (auto j : node[i]) {
      if (vis[j] == 1) {
        cnt++;
      }
    }
    if (cnt >= 2) ans2 = ans2 - cnt + 1;
  }
  cout << ans << " " << ans2;
}

```