#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-7;
const int N = 200010;
const int M = 200010;
const int P = 1000000007;
const int Inf = 1000000007;
const int Fni = -1000000007;
inline int qr() {
  int r = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    r = (r << 1) + (r << 3) + c - '0', c = getchar();
  }
  return r * f;
}
template <typename T>
void debug(T x, char c = '\n') {
  cerr << x << c;
}
template <typename T>
void debug(T *const tp, int minn, int maxn, char v = ' ', char c = '\n') {
  for (int i = minn; i <= maxn; ++i) debug(tp[i], v);
  cerr << c;
}
template <typename T>
inline void chkmin(T &x, T y) {
  x = x > y ? y : x;
}
template <typename T>
inline void chkmax(T &x, T y) {
  x = x < y ? y : x;
}
template <typename T>
inline void add(T &x, long long y, int mod = P) {
  x += y;
  x = x >= mod ? x - mod : x;
}
template <typename T>
inline void dec(T &x, long long y, int mod = P) {
  x -= y;
  x = x < 0 ? x + mod : x;
}
template <typename T>
inline T addn(T x, long long y, int mod = P) {
  x += y;
  return (x = x > mod ? x - mod : x);
}
template <typename T>
inline T decn(T x, long long y, int mod = P) {
  x -= y;
  return (x = x < 0 ? x + mod : x);
}
int ans;
int res1;
int res2;
int root;
int T, n;
int buc[N];
int deg[N];
int dep[N];
vector<int> E[N];
void dfs(int x, int y) {
  int qwq = 0;
  dep[x] = dep[y] + 1;
  for (auto k : E[x])
    if (k != y) {
      dfs(k, x);
      qwq += (deg[k] <= 1);
    }
  if (qwq > 1) res2 -= qwq - 1;
  if (deg[x] == 1 && x != root) ans = ans & (dep[x] & 1), buc[dep[x]]++;
}
int main() {
  cin >> n;
  int x, y;
  ans = 1;
  for (int i = 1; i < n; ++i) {
    x = qr(), y = qr();
    deg[x]++, deg[y]++;
    E[x].push_back(y), E[y].push_back(x);
  }
  for (int i = 1; i <= n; ++i)
    if (deg[i] <= 1) root = i;
  dfs(root, 0);
  res2 += n - 1;
  res1 = 1 + 2 * (!ans);
  if (buc[3]) res2--;
  cout << res1 << " " << res2 << '\n';
  return 0;
}

### Prompt

Your task is to create 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 double eps = 1e-7;
const int N = 200010;
const int M = 200010;
const int P = 1000000007;
const int Inf = 1000000007;
const int Fni = -1000000007;
inline int qr() {
  int r = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    r = (r << 1) + (r << 3) + c - '0', c = getchar();
  }
  return r * f;
}
template <typename T>
void debug(T x, char c = '\n') {
  cerr << x << c;
}
template <typename T>
void debug(T *const tp, int minn, int maxn, char v = ' ', char c = '\n') {
  for (int i = minn; i <= maxn; ++i) debug(tp[i], v);
  cerr << c;
}
template <typename T>
inline void chkmin(T &x, T y) {
  x = x > y ? y : x;
}
template <typename T>
inline void chkmax(T &x, T y) {
  x = x < y ? y : x;
}
template <typename T>
inline void add(T &x, long long y, int mod = P) {
  x += y;
  x = x >= mod ? x - mod : x;
}
template <typename T>
inline void dec(T &x, long long y, int mod = P) {
  x -= y;
  x = x < 0 ? x + mod : x;
}
template <typename T>
inline T addn(T x, long long y, int mod = P) {
  x += y;
  return (x = x > mod ? x - mod : x);
}
template <typename T>
inline T decn(T x, long long y, int mod = P) {
  x -= y;
  return (x = x < 0 ? x + mod : x);
}
int ans;
int res1;
int res2;
int root;
int T, n;
int buc[N];
int deg[N];
int dep[N];
vector<int> E[N];
void dfs(int x, int y) {
  int qwq = 0;
  dep[x] = dep[y] + 1;
  for (auto k : E[x])
    if (k != y) {
      dfs(k, x);
      qwq += (deg[k] <= 1);
    }
  if (qwq > 1) res2 -= qwq - 1;
  if (deg[x] == 1 && x != root) ans = ans & (dep[x] & 1), buc[dep[x]]++;
}
int main() {
  cin >> n;
  int x, y;
  ans = 1;
  for (int i = 1; i < n; ++i) {
    x = qr(), y = qr();
    deg[x]++, deg[y]++;
    E[x].push_back(y), E[y].push_back(x);
  }
  for (int i = 1; i <= n; ++i)
    if (deg[i] <= 1) root = i;
  dfs(root, 0);
  res2 += n - 1;
  res1 = 1 + 2 * (!ans);
  if (buc[3]) res2--;
  cout << res1 << " " << res2 << '\n';
  return 0;
}

```