#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= f;
}
template <typename T, typename... Args>
void read(T &first, Args &...args) {
  read(first);
  read(args...);
}
template <typename T>
void write(T arg) {
  T x = arg;
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
template <typename T, typename... Ts>
void write(T arg, Ts... args) {
  write(arg);
  if (sizeof...(args) != 0) {
    putchar(' ');
    write(args...);
  }
}
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
int n, m;
const int N = 4e5 + 5;
vector<int> edge[N];
int deg[N], dep[N], mark[N], pa[N];
vector<int> leaf;
void dfs(int u, int fa) {
  for (auto v : edge[u]) {
    if (v != fa) {
      pa[v] = u;
      dep[v] = dep[u] + 1;
      dfs(v, u);
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    edge[u].push_back(v);
    edge[v].push_back(u);
    deg[u]++, deg[v]++;
  }
  int rt = -1;
  for (int i = 1; i <= n; i++) {
    if (deg[i] != 1)
      rt = i;
    else
      leaf.push_back(i), mark[i] = 1;
  }
  dfs(rt, 0);
  int ok = 1;
  for (int i = 1; i < leaf.size(); i++) {
    if (dep[leaf[i]] % 2 != dep[leaf[i - 1]] % 2) ok = 0;
  }
  if (ok)
    cout << 1 << ' ';
  else
    cout << 3 << ' ';
  vector<int> a;
  for (auto v : leaf) {
    if (pa[v]) a.push_back(pa[v]);
  }
  sort(a.begin(), a.end());
  int maxn = n - 1;
  for (int i = 1; i < a.size(); i++) {
    if (a[i] == a[i - 1]) maxn--;
  }
  cout << maxn << endl;
}

### Prompt

Please create a solution in CPP 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;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= f;
}
template <typename T, typename... Args>
void read(T &first, Args &...args) {
  read(first);
  read(args...);
}
template <typename T>
void write(T arg) {
  T x = arg;
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
template <typename T, typename... Ts>
void write(T arg, Ts... args) {
  write(arg);
  if (sizeof...(args) != 0) {
    putchar(' ');
    write(args...);
  }
}
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
int n, m;
const int N = 4e5 + 5;
vector<int> edge[N];
int deg[N], dep[N], mark[N], pa[N];
vector<int> leaf;
void dfs(int u, int fa) {
  for (auto v : edge[u]) {
    if (v != fa) {
      pa[v] = u;
      dep[v] = dep[u] + 1;
      dfs(v, u);
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    edge[u].push_back(v);
    edge[v].push_back(u);
    deg[u]++, deg[v]++;
  }
  int rt = -1;
  for (int i = 1; i <= n; i++) {
    if (deg[i] != 1)
      rt = i;
    else
      leaf.push_back(i), mark[i] = 1;
  }
  dfs(rt, 0);
  int ok = 1;
  for (int i = 1; i < leaf.size(); i++) {
    if (dep[leaf[i]] % 2 != dep[leaf[i - 1]] % 2) ok = 0;
  }
  if (ok)
    cout << 1 << ' ';
  else
    cout << 3 << ' ';
  vector<int> a;
  for (auto v : leaf) {
    if (pa[v]) a.push_back(pa[v]);
  }
  sort(a.begin(), a.end());
  int maxn = n - 1;
  for (int i = 1; i < a.size(); i++) {
    if (a[i] == a[i - 1]) maxn--;
  }
  cout << maxn << endl;
}

```