#include <bits/stdc++.h>
using namespace std;
inline long long gcd(long long a, long long b) {
  return b > 0 ? gcd(b, a % b) : a;
}
inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
inline long long qpow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) {
      ans *= a;
    }
    a *= a;
    n >>= 1;
  }
  return ans;
}
inline long long _qpow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) {
      ans *= a % 1000000007;
      ans %= 1000000007;
    }
    a *= a;
    a %= 1000000007;
    n >>= 1;
  }
  return ans % 1000000007;
}
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') {
      f = -1;
    }
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const long long maxnum = 1e6 + 5;
vector<long long> G[maxnum];
long long d[maxnum];
long long cot[maxnum];
void _dfs(long long u, long long fa) {
  d[u] = d[fa] + 1;
  for (long long v : G[u]) {
    if (v == fa) continue;
    _dfs(v, u);
  }
}
vector<long long> lv[maxnum];
long long dp[maxnum];
bool vis[maxnum];
long long dfs(long long u, long long fa) {
  for (long long v : G[u]) {
    if (v == fa) continue;
    if (vis[v]) continue;
    dp[u] += dfs(v, u) + 1;
  }
  return dp[u];
}
signed main() {
  long long n = read();
  long long u, v;
  for (long long i = 1; i <= n - 1; i++) {
    u = read(), v = read();
    G[u].push_back(v);
    G[v].push_back(u);
    cot[u]++, cot[v]++;
  }
  long long root = 0;
  for (long long i = 1; i <= n; i++) {
    if (cot[i] > 1) {
      root = i;
      break;
    }
  }
  _dfs(root, 0);
  set<long long> st;
  for (long long i = 1; i <= n; i++) {
    if (cot[i] == 1) {
      st.insert(d[i] % 2);
    }
  }
  if (st.size() == 1) {
    cout << "1 ";
  } else {
    cout << "3 ";
  }
  for (long long i = 1; i <= n; i++) {
    if (cot[i] == 1) {
      u = i, v = G[i].back();
      lv[v].push_back(u);
    }
  }
  long long res = 0;
  for (long long i = 1; i <= n; i++) {
    if (lv[i].size() >= 2) {
      res += lv[i].size() - 1;
    }
  }
  cout << n - 1 - res << endl;
  return 0;
}

### 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;
inline long long gcd(long long a, long long b) {
  return b > 0 ? gcd(b, a % b) : a;
}
inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
inline long long qpow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) {
      ans *= a;
    }
    a *= a;
    n >>= 1;
  }
  return ans;
}
inline long long _qpow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) {
      ans *= a % 1000000007;
      ans %= 1000000007;
    }
    a *= a;
    a %= 1000000007;
    n >>= 1;
  }
  return ans % 1000000007;
}
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') {
      f = -1;
    }
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const long long maxnum = 1e6 + 5;
vector<long long> G[maxnum];
long long d[maxnum];
long long cot[maxnum];
void _dfs(long long u, long long fa) {
  d[u] = d[fa] + 1;
  for (long long v : G[u]) {
    if (v == fa) continue;
    _dfs(v, u);
  }
}
vector<long long> lv[maxnum];
long long dp[maxnum];
bool vis[maxnum];
long long dfs(long long u, long long fa) {
  for (long long v : G[u]) {
    if (v == fa) continue;
    if (vis[v]) continue;
    dp[u] += dfs(v, u) + 1;
  }
  return dp[u];
}
signed main() {
  long long n = read();
  long long u, v;
  for (long long i = 1; i <= n - 1; i++) {
    u = read(), v = read();
    G[u].push_back(v);
    G[v].push_back(u);
    cot[u]++, cot[v]++;
  }
  long long root = 0;
  for (long long i = 1; i <= n; i++) {
    if (cot[i] > 1) {
      root = i;
      break;
    }
  }
  _dfs(root, 0);
  set<long long> st;
  for (long long i = 1; i <= n; i++) {
    if (cot[i] == 1) {
      st.insert(d[i] % 2);
    }
  }
  if (st.size() == 1) {
    cout << "1 ";
  } else {
    cout << "3 ";
  }
  for (long long i = 1; i <= n; i++) {
    if (cot[i] == 1) {
      u = i, v = G[i].back();
      lv[v].push_back(u);
    }
  }
  long long res = 0;
  for (long long i = 1; i <= n; i++) {
    if (lv[i].size() >= 2) {
      res += lv[i].size() - 1;
    }
  }
  cout << n - 1 - res << endl;
  return 0;
}

```