#include <bits/stdc++.h>
const int MaxN = 100000;
struct Graph {
  int cnte;
  int Head[MaxN + 5], To[MaxN * 2 + 5], Next[MaxN * 2 + 5];
  inline void addEdge(int from, int to) {
    cnte++;
    To[cnte] = to;
    Next[cnte] = Head[from];
    Head[from] = cnte;
  }
};
int N;
int Dep[MaxN + 5], Deg[MaxN + 5], HaveLeaf[MaxN + 5];
Graph G;
void init() {
  scanf("%d", &N);
  for (int i = 1; i < N; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    G.addEdge(u, v);
    G.addEdge(v, u);
    Deg[u]++, Deg[v]++;
  }
}
void dfs(int u, int f) {
  for (int i = G.Head[u]; i; i = G.Next[i]) {
    int v = G.To[i];
    if (v == f) continue;
    Dep[v] = Dep[u] + 1;
    dfs(v, u);
  }
}
void solve() {
  dfs(1, 0);
  bool haveOddLeaf = false, haveEvenLeaf = false;
  for (int i = 1; i <= N; ++i) {
    if (Deg[i] != 1) continue;
    if (Dep[i] & 1)
      haveOddLeaf = true;
    else
      haveEvenLeaf = true;
  }
  if (haveOddLeaf && haveEvenLeaf)
    printf("%d ", 3);
  else
    printf("%d ", 1);
  int locked = 0;
  for (int i = 1; i <= N; ++i) {
    if (Deg[i] != 1) continue;
    int v = G.To[G.Head[i]];
    HaveLeaf[v]++;
  }
  for (int i = 1; i <= N; ++i)
    if (HaveLeaf[i] != 0) locked += HaveLeaf[i] - 1;
  printf("%d\n", N - 1 - locked);
}
int main() {
  init();
  solve();
  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>
const int MaxN = 100000;
struct Graph {
  int cnte;
  int Head[MaxN + 5], To[MaxN * 2 + 5], Next[MaxN * 2 + 5];
  inline void addEdge(int from, int to) {
    cnte++;
    To[cnte] = to;
    Next[cnte] = Head[from];
    Head[from] = cnte;
  }
};
int N;
int Dep[MaxN + 5], Deg[MaxN + 5], HaveLeaf[MaxN + 5];
Graph G;
void init() {
  scanf("%d", &N);
  for (int i = 1; i < N; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    G.addEdge(u, v);
    G.addEdge(v, u);
    Deg[u]++, Deg[v]++;
  }
}
void dfs(int u, int f) {
  for (int i = G.Head[u]; i; i = G.Next[i]) {
    int v = G.To[i];
    if (v == f) continue;
    Dep[v] = Dep[u] + 1;
    dfs(v, u);
  }
}
void solve() {
  dfs(1, 0);
  bool haveOddLeaf = false, haveEvenLeaf = false;
  for (int i = 1; i <= N; ++i) {
    if (Deg[i] != 1) continue;
    if (Dep[i] & 1)
      haveOddLeaf = true;
    else
      haveEvenLeaf = true;
  }
  if (haveOddLeaf && haveEvenLeaf)
    printf("%d ", 3);
  else
    printf("%d ", 1);
  int locked = 0;
  for (int i = 1; i <= N; ++i) {
    if (Deg[i] != 1) continue;
    int v = G.To[G.Head[i]];
    HaveLeaf[v]++;
  }
  for (int i = 1; i <= N; ++i)
    if (HaveLeaf[i] != 0) locked += HaveLeaf[i] - 1;
  printf("%d\n", N - 1 - locked);
}
int main() {
  init();
  solve();
  return 0;
}

```