#include <bits/stdc++.h>
using namespace std;
void read() { return; }
void print() { return; }
template <typename T1, typename... T2>
void read(T1 &x, T2 &...args) {
  ((cin >> x), read(args...));
}
template <typename T1, typename... T2>
void print(T1 x, T2... args) {
  ((cout << x << ' '), print(args...));
}
template <typename T>
void readArr(T &arr, int x, int y) {
  for (int i = (int)x; i < (int)y; ++i) cin >> arr[i];
}
template <typename T>
void printArr(T &arr, int x, int y) {
  for (int i = (int)x; i < (int)y; ++i) cout << arr[i] << ' ';
}
const int N = 1e6;
const int INF = 1e9;
const int MOD = 1e9 + 7;
int countBits(long long number) { return (int)log2(number) + 1; }
long long gcd(long long a, long long b) {
  return (b == 0) ? (a) : (gcd(b, a %= b));
}
long long add(long long x, long long y) { return (x + y) % MOD; }
long long sub(long long x, long long y) { return (x - y + MOD) % MOD; }
long long mul(long long x, long long y) { return (x * 1ll * y) % MOD; }
long long inv(long long p, long long q) {
  long long expo = MOD - 2;
  while (expo) {
    if (expo & 1) p = mul(p, q);
    q = mul(q, q), expo >>= 1;
  }
  return p;
}
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0) {
    long long tmp = power(x, y / 2);
    return mul(tmp, tmp);
  }
  return mul(x, power(x, y - 1));
}
vector<int> g[N];
int cnt = 0;
void dfs(int s, int e = 0, int h = 0) {
  int flag = 1;
  for (int v : g[s])
    if (v != e) {
      dfs(v, s, h + 1);
      flag = 0;
    }
  if (flag && (h & 1)) cnt++;
}
void solveTestCase() {
  int n;
  read(n);
  int u, v;
  for (int i = (int)0; i < (int)n - 1; ++i) {
    read(u, v);
    g[u].emplace_back(v), g[v].emplace_back(u);
  }
  int leaf = -1, lfCnt = 0;
  set<int> st;
  for (int i = (int)1; i < (int)n + 1; ++i)
    if (g[i].size() == 1) leaf = i, lfCnt++, st.insert(g[i][0]);
  dfs(leaf);
  int mn = (cnt) ? 3 : 1;
  int mx = n - 1 - lfCnt + st.size();
  print(mn, mx);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int t = 1;
  while (t--) solveTestCase();
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
void read() { return; }
void print() { return; }
template <typename T1, typename... T2>
void read(T1 &x, T2 &...args) {
  ((cin >> x), read(args...));
}
template <typename T1, typename... T2>
void print(T1 x, T2... args) {
  ((cout << x << ' '), print(args...));
}
template <typename T>
void readArr(T &arr, int x, int y) {
  for (int i = (int)x; i < (int)y; ++i) cin >> arr[i];
}
template <typename T>
void printArr(T &arr, int x, int y) {
  for (int i = (int)x; i < (int)y; ++i) cout << arr[i] << ' ';
}
const int N = 1e6;
const int INF = 1e9;
const int MOD = 1e9 + 7;
int countBits(long long number) { return (int)log2(number) + 1; }
long long gcd(long long a, long long b) {
  return (b == 0) ? (a) : (gcd(b, a %= b));
}
long long add(long long x, long long y) { return (x + y) % MOD; }
long long sub(long long x, long long y) { return (x - y + MOD) % MOD; }
long long mul(long long x, long long y) { return (x * 1ll * y) % MOD; }
long long inv(long long p, long long q) {
  long long expo = MOD - 2;
  while (expo) {
    if (expo & 1) p = mul(p, q);
    q = mul(q, q), expo >>= 1;
  }
  return p;
}
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0) {
    long long tmp = power(x, y / 2);
    return mul(tmp, tmp);
  }
  return mul(x, power(x, y - 1));
}
vector<int> g[N];
int cnt = 0;
void dfs(int s, int e = 0, int h = 0) {
  int flag = 1;
  for (int v : g[s])
    if (v != e) {
      dfs(v, s, h + 1);
      flag = 0;
    }
  if (flag && (h & 1)) cnt++;
}
void solveTestCase() {
  int n;
  read(n);
  int u, v;
  for (int i = (int)0; i < (int)n - 1; ++i) {
    read(u, v);
    g[u].emplace_back(v), g[v].emplace_back(u);
  }
  int leaf = -1, lfCnt = 0;
  set<int> st;
  for (int i = (int)1; i < (int)n + 1; ++i)
    if (g[i].size() == 1) leaf = i, lfCnt++, st.insert(g[i][0]);
  dfs(leaf);
  int mn = (cnt) ? 3 : 1;
  int mx = n - 1 - lfCnt + st.size();
  print(mn, mx);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int t = 1;
  while (t--) solveTestCase();
  return 0;
}

```