#include <bits/stdc++.h>
using namespace std;
const int N = 1e2 + 5;
const long long mod = 1e9 + 7;
long long powermodm(long long x, long long n, long long M) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
long long power(long long x, long long y) {
  long long z = 1;
  while (y) {
    if (y % 2 == 1) z = (z * x);
    y /= 2;
    x = (x * x);
  }
  return z;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) {
  return (max(a, b) / gcd(a, b)) * min(a, b);
}
set<long long> s;
long long edges, leaves, extra;
vector<long long> edg[100001];
void dfs(long long u, long long p, long long lev) {
  bool flag = false;
  for (long long i = 0; i < edg[u].size(); i++) {
    if (edg[u][i] == p) continue;
    dfs(edg[u][i], u, lev + 1);
    edges++;
    if (edg[edg[u][i]].size() == 1) flag = true;
  }
  if (flag) extra++;
  if (edg[u].size() == 1) {
    s.insert(lev % 2);
    leaves++;
  }
}
int main() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    edg[x].push_back(y);
    edg[y].push_back(x);
  }
  long long root;
  for (long long i = 1; i <= n; i++) {
    if (edg[i].size() > 1) root = i;
  }
  dfs(root, 0, 0);
  long long ans1, ans2;
  ans2 = edges - leaves + extra;
  if (s.size() == 1)
    ans1 = 1;
  else
    ans1 = 3;
  cout << ans1 << " " << ans2 << endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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 int N = 1e2 + 5;
const long long mod = 1e9 + 7;
long long powermodm(long long x, long long n, long long M) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
long long power(long long x, long long y) {
  long long z = 1;
  while (y) {
    if (y % 2 == 1) z = (z * x);
    y /= 2;
    x = (x * x);
  }
  return z;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) {
  return (max(a, b) / gcd(a, b)) * min(a, b);
}
set<long long> s;
long long edges, leaves, extra;
vector<long long> edg[100001];
void dfs(long long u, long long p, long long lev) {
  bool flag = false;
  for (long long i = 0; i < edg[u].size(); i++) {
    if (edg[u][i] == p) continue;
    dfs(edg[u][i], u, lev + 1);
    edges++;
    if (edg[edg[u][i]].size() == 1) flag = true;
  }
  if (flag) extra++;
  if (edg[u].size() == 1) {
    s.insert(lev % 2);
    leaves++;
  }
}
int main() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    edg[x].push_back(y);
    edg[y].push_back(x);
  }
  long long root;
  for (long long i = 1; i <= n; i++) {
    if (edg[i].size() > 1) root = i;
  }
  dfs(root, 0, 0);
  long long ans1, ans2;
  ans2 = edges - leaves + extra;
  if (s.size() == 1)
    ans1 = 1;
  else
    ans1 = 3;
  cout << ans1 << " " << ans2 << endl;
  return 0;
}

```