#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const int maxn = 2e5 + 5;
const int maxm = 1e6 + 5;
int n;
int head[maxn], point[maxn], nxt[maxn], ind[maxn], size;
int dep[maxn], lef[maxn], sub = 0;
void init() {
  size = 0;
  memset(head, -1, sizeof(head));
}
void add(int a, int b) {
  point[size] = b;
  nxt[size] = head[a];
  head[a] = size++;
  ind[b]++;
}
void dfs(int s, int f, int d) {
  dep[s] = d;
  int res = 0;
  for (int i = head[s]; ~i; i = nxt[i]) {
    int j = point[i];
    if (j == f) continue;
    if (lef[j]) res++;
    dfs(j, s, d + 1);
  }
  if (res) sub += res - 1;
}
void solve() {
  init();
  for (int i = 1; i < n; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    add(a, b);
    add(b, a);
  }
  int root = 0;
  for (int i = 1; i <= n; ++i) {
    if (ind[i] > 1)
      root = i;
    else
      lef[i] = 1;
  }
  dfs(root, -1, 1);
  int mn = 0, mx = n - 1 - sub;
  int odd = 0, even = 0;
  for (int i = 1; i <= n; ++i) {
    if (!lef[i]) continue;
    if (dep[i] & 1)
      odd = 1;
    else
      even = 1;
  }
  if (even && odd)
    mn = 3;
  else
    mn = 1;
  printf("%d %d\n", mn, mx);
}
int main() {
  scanf("%d", &n);
  solve();
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const int maxn = 2e5 + 5;
const int maxm = 1e6 + 5;
int n;
int head[maxn], point[maxn], nxt[maxn], ind[maxn], size;
int dep[maxn], lef[maxn], sub = 0;
void init() {
  size = 0;
  memset(head, -1, sizeof(head));
}
void add(int a, int b) {
  point[size] = b;
  nxt[size] = head[a];
  head[a] = size++;
  ind[b]++;
}
void dfs(int s, int f, int d) {
  dep[s] = d;
  int res = 0;
  for (int i = head[s]; ~i; i = nxt[i]) {
    int j = point[i];
    if (j == f) continue;
    if (lef[j]) res++;
    dfs(j, s, d + 1);
  }
  if (res) sub += res - 1;
}
void solve() {
  init();
  for (int i = 1; i < n; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    add(a, b);
    add(b, a);
  }
  int root = 0;
  for (int i = 1; i <= n; ++i) {
    if (ind[i] > 1)
      root = i;
    else
      lef[i] = 1;
  }
  dfs(root, -1, 1);
  int mn = 0, mx = n - 1 - sub;
  int odd = 0, even = 0;
  for (int i = 1; i <= n; ++i) {
    if (!lef[i]) continue;
    if (dep[i] & 1)
      odd = 1;
    else
      even = 1;
  }
  if (even && odd)
    mn = 3;
  else
    mn = 1;
  printf("%d %d\n", mn, mx);
}
int main() {
  scanf("%d", &n);
  solve();
  return 0;
}

```