#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
const int N = 1e5 + 10;
int n, d[N], deg[N], f[N];
bool fl;
vector<int> ad[N];
set<int> s;
void dfs(int v, int par) {
  for (auto u : ad[v]) {
    if (u == par) continue;
    d[u] = d[v] + 1;
    dfs(u, v);
  }
}
void Dfs(int v, int par) {
  if (ad[v].size() == 1) f[par]++, s.insert(par);
  for (auto u : ad[v]) {
    if (u == par) continue;
    Dfs(u, v);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    ad[x].push_back(y);
    ad[y].push_back(x);
    deg[x]++;
    deg[y]++;
  }
  int a = 0, mx = 0;
  dfs(1, 0);
  for (int i = 1; i <= n; i++) {
    if (d[i] > mx) {
      mx = d[i];
      a = i;
    }
  }
  for (int i = 1; i <= n; i++) d[i] = 0;
  dfs(a, 0);
  for (int i = 1; i <= n; i++) {
    if (deg[i] > 1) {
      Dfs(i, 0);
      break;
    }
  }
  int di = *max_element(d, d + 1 + n);
  for (int i = 1; i <= n; i++)
    if ((d[i] & 1) && ad[i].size() == 1) fl = 1;
  if (fl || (di & 1))
    printf("%d ", 3);
  else
    printf("%d ", 1);
  int ans = 0, tot = 0;
  for (int i = 1; i <= n; i++) ans += (f[i] > 0), tot += f[i];
  ans += (n - 1) - tot;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate 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;
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
const int N = 1e5 + 10;
int n, d[N], deg[N], f[N];
bool fl;
vector<int> ad[N];
set<int> s;
void dfs(int v, int par) {
  for (auto u : ad[v]) {
    if (u == par) continue;
    d[u] = d[v] + 1;
    dfs(u, v);
  }
}
void Dfs(int v, int par) {
  if (ad[v].size() == 1) f[par]++, s.insert(par);
  for (auto u : ad[v]) {
    if (u == par) continue;
    Dfs(u, v);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    ad[x].push_back(y);
    ad[y].push_back(x);
    deg[x]++;
    deg[y]++;
  }
  int a = 0, mx = 0;
  dfs(1, 0);
  for (int i = 1; i <= n; i++) {
    if (d[i] > mx) {
      mx = d[i];
      a = i;
    }
  }
  for (int i = 1; i <= n; i++) d[i] = 0;
  dfs(a, 0);
  for (int i = 1; i <= n; i++) {
    if (deg[i] > 1) {
      Dfs(i, 0);
      break;
    }
  }
  int di = *max_element(d, d + 1 + n);
  for (int i = 1; i <= n; i++)
    if ((d[i] & 1) && ad[i].size() == 1) fl = 1;
  if (fl || (di & 1))
    printf("%d ", 3);
  else
    printf("%d ", 1);
  int ans = 0, tot = 0;
  for (int i = 1; i <= n; i++) ans += (f[i] > 0), tot += f[i];
  ans += (n - 1) - tot;
  printf("%d\n", ans);
  return 0;
}

```