#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:10000000000")
using namespace std;
const int INF = 1000000007;
const long long INF2 = 1000000000000000007;
const long long MOD = 1172113087;
const long long MOD1 = 1864463441;
const int SIZE = 200010;
const int base = 53;
const double eps = 1e-12;
const double pi = acos(-1);
long long n, v, v1, mx, mx1, mi1, b, a, a1, minres, maxres, b1, mi;
vector<long long> g[SIZE];
long long dd[SIZE];
bool used[SIZE];
vector<long long> toptop;
void dfs(int v, int depth = 0) {
  if (used[v]) return;
  used[v] = 1;
  dd[v] = depth;
  bool zhopa = 0;
  ;
  for (auto& i : g[v]) {
    if (g[i].size() == 1) {
      maxres--;
      zhopa = 1;
    }
    dfs(i, depth + 1);
  }
  if (zhopa) maxres++;
  toptop.push_back(v);
}
vector<long long> Left, Right;
bool fl;
int cheta, chetb, necheta, nechetb;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; ++i) {
    cin >> v >> v1;
    v--;
    v1--;
    g[v].push_back(v1);
    g[v1].push_back(v);
  }
  maxres = n - 1;
  dfs(0);
  fl = 0;
  for (int i = 0; i < toptop.size(); ++i) {
    if (toptop[i] == 0) {
      fl = 1;
    }
    if (g[toptop[i]].size() == 1) {
      if (!fl) {
        Left.push_back(dd[toptop[i]]);
        if (dd[toptop[i]] % 2)
          cheta++;
        else
          necheta++;
      } else {
        Right.push_back(dd[toptop[i]]);
        if (dd[toptop[i]] % 2)
          cheta++;
        else
          necheta++;
      }
    }
  }
  sort(Left.begin(), Left.end());
  sort(Right.begin(), Right.end());
  a = INF2;
  a1 = INF2;
  if (Left.size()) {
    mi += *Left.begin();
    if (Left.size() > 1) {
      a = Left[1] - Left[0];
    }
  }
  if (Right.size()) {
    mi += *Right.begin();
    if (Right.size() > 1) {
      a1 = Right[1] - Right[0];
    }
  }
  if (Left.empty() || Right.empty()) {
    mi = INF2;
  }
  minres = 1;
  mi = min(mi, min(a + 1, a1 + 1));
  if (necheta && cheta) minres = 3;
  cout << minres << " " << maxres;
  return 0;
}

### Prompt

Please formulate 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>
#pragma comment(linker, "/STACK:10000000000")
using namespace std;
const int INF = 1000000007;
const long long INF2 = 1000000000000000007;
const long long MOD = 1172113087;
const long long MOD1 = 1864463441;
const int SIZE = 200010;
const int base = 53;
const double eps = 1e-12;
const double pi = acos(-1);
long long n, v, v1, mx, mx1, mi1, b, a, a1, minres, maxres, b1, mi;
vector<long long> g[SIZE];
long long dd[SIZE];
bool used[SIZE];
vector<long long> toptop;
void dfs(int v, int depth = 0) {
  if (used[v]) return;
  used[v] = 1;
  dd[v] = depth;
  bool zhopa = 0;
  ;
  for (auto& i : g[v]) {
    if (g[i].size() == 1) {
      maxres--;
      zhopa = 1;
    }
    dfs(i, depth + 1);
  }
  if (zhopa) maxres++;
  toptop.push_back(v);
}
vector<long long> Left, Right;
bool fl;
int cheta, chetb, necheta, nechetb;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; ++i) {
    cin >> v >> v1;
    v--;
    v1--;
    g[v].push_back(v1);
    g[v1].push_back(v);
  }
  maxres = n - 1;
  dfs(0);
  fl = 0;
  for (int i = 0; i < toptop.size(); ++i) {
    if (toptop[i] == 0) {
      fl = 1;
    }
    if (g[toptop[i]].size() == 1) {
      if (!fl) {
        Left.push_back(dd[toptop[i]]);
        if (dd[toptop[i]] % 2)
          cheta++;
        else
          necheta++;
      } else {
        Right.push_back(dd[toptop[i]]);
        if (dd[toptop[i]] % 2)
          cheta++;
        else
          necheta++;
      }
    }
  }
  sort(Left.begin(), Left.end());
  sort(Right.begin(), Right.end());
  a = INF2;
  a1 = INF2;
  if (Left.size()) {
    mi += *Left.begin();
    if (Left.size() > 1) {
      a = Left[1] - Left[0];
    }
  }
  if (Right.size()) {
    mi += *Right.begin();
    if (Right.size() > 1) {
      a1 = Right[1] - Right[0];
    }
  }
  if (Left.empty() || Right.empty()) {
    mi = INF2;
  }
  minres = 1;
  mi = min(mi, min(a + 1, a1 + 1));
  if (necheta && cheta) minres = 3;
  cout << minres << " " << maxres;
  return 0;
}

```