#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using lpair = pair<ll, ll>;
const ll MOD = 1e9 + 7;
const ll INF = 1e18;
void print() {}
template <class H, class... T>
void print(H &&h, T &&...t) {
  cout << h << " \n"[sizeof...(t) == 0];
  print(forward<T>(t)...);
}
ll depth[100010] = {};
vector<ll> tree[100010];
void dfs(ll cur, ll par, ll d) {
  depth[cur] = d;
  for (auto &e : tree[cur]) {
    if (e == par) continue;
    dfs(e, cur, d + 1);
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  ll N;
  cin >> N;
  ll a[100010], b[100010];
  for (ll i = (0); i < (N - 1); i++) {
    cin >> a[i] >> b[i];
    a[i]--;
    b[i]--;
    tree[a[i]].push_back(b[i]);
    tree[b[i]].push_back(a[i]);
  }
  dfs(0, -1, 0);
  ll max_val = N - 1, min_val = 0;
  ll leaf_cnt[100010] = {};
  vector<ll> p;
  for (ll i = (0); i < (N); i++) {
    if (tree[i].size() == 1) {
      leaf_cnt[tree[i][0]]++;
      p.push_back(tree[i][0]);
    }
  }
  p.erase(unique(p.begin(), p.end()), p.end());
  ll sz = p.size();
  ll num = 0;
  for (auto &e : p) {
    if (depth[e] % 2 == 0) {
      num++;
    }
  }
  if (num == sz || num == 0) {
    min_val = 1;
  } else {
    min_val = 3;
  }
  for (ll i = (0); i < (N); i++) {
    if (leaf_cnt[i] > 1) {
      max_val -= leaf_cnt[i] - 1;
    }
  }
  print(min_val, max_val);
}

### 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;
using ll = long long;
using lpair = pair<ll, ll>;
const ll MOD = 1e9 + 7;
const ll INF = 1e18;
void print() {}
template <class H, class... T>
void print(H &&h, T &&...t) {
  cout << h << " \n"[sizeof...(t) == 0];
  print(forward<T>(t)...);
}
ll depth[100010] = {};
vector<ll> tree[100010];
void dfs(ll cur, ll par, ll d) {
  depth[cur] = d;
  for (auto &e : tree[cur]) {
    if (e == par) continue;
    dfs(e, cur, d + 1);
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  ll N;
  cin >> N;
  ll a[100010], b[100010];
  for (ll i = (0); i < (N - 1); i++) {
    cin >> a[i] >> b[i];
    a[i]--;
    b[i]--;
    tree[a[i]].push_back(b[i]);
    tree[b[i]].push_back(a[i]);
  }
  dfs(0, -1, 0);
  ll max_val = N - 1, min_val = 0;
  ll leaf_cnt[100010] = {};
  vector<ll> p;
  for (ll i = (0); i < (N); i++) {
    if (tree[i].size() == 1) {
      leaf_cnt[tree[i][0]]++;
      p.push_back(tree[i][0]);
    }
  }
  p.erase(unique(p.begin(), p.end()), p.end());
  ll sz = p.size();
  ll num = 0;
  for (auto &e : p) {
    if (depth[e] % 2 == 0) {
      num++;
    }
  }
  if (num == sz || num == 0) {
    min_val = 1;
  } else {
    min_val = 3;
  }
  for (ll i = (0); i < (N); i++) {
    if (leaf_cnt[i] > 1) {
      max_val -= leaf_cnt[i] - 1;
    }
  }
  print(min_val, max_val);
}

```