#include <bits/stdc++.h>
using namespace std;
const int INF = 2147483647;
const int INF2 = 0x3f3f3f3f;
const long long INF64 = 0x3f3f3f3f3f3f3f3f;
const double PI = 3.14159265358979323846;
const long long MOD = 1e9 + 7;
const long long TARGET = (1LL << 32) - 1LL;
template <typename T>
T Max(T val1, T val2) {
  return val1 > val2 ? val1 : val2;
}
template <typename T, typename... Args>
T Max(T val, Args... arg) {
  T result = Max(arg...);
  return Max(val, result);
}
template <typename T>
inline T read() {
  T X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  return w ? -X : X;
}
const int MAXN = 500005;
int n, m, k;
int deg[MAXN];
vector<int> G[MAXN];
int dep[MAXN];
int ans;
void dfs(int u, int p) {
  int cnt = 0;
  for (auto v : G[u]) {
    if (v == p) continue;
    dep[v] = dep[u] + 1;
    dfs(v, u);
    if (deg[v] == 1) cnt++;
  }
  ans -= max(0, cnt - 1);
}
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v), G[v].push_back(u);
    deg[u]++, deg[v]++;
  }
  int rt = 1;
  for (int i = 1; i <= n; i++) {
    if (deg[i] != 1) {
      rt = i;
      break;
    }
  }
  ans = n - 1;
  dfs(rt, rt);
  int odd = -1;
  bool flag = false;
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 1) {
      if (odd < 0)
        odd = dep[i] & 1;
      else if ((dep[i] & 1) != odd) {
        flag = true;
      }
    }
  }
  if (flag) {
    printf("3 %d\n", ans);
  } else {
    printf("1 %d\n", ans);
  }
}
int main() {
  solve();
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 INF = 2147483647;
const int INF2 = 0x3f3f3f3f;
const long long INF64 = 0x3f3f3f3f3f3f3f3f;
const double PI = 3.14159265358979323846;
const long long MOD = 1e9 + 7;
const long long TARGET = (1LL << 32) - 1LL;
template <typename T>
T Max(T val1, T val2) {
  return val1 > val2 ? val1 : val2;
}
template <typename T, typename... Args>
T Max(T val, Args... arg) {
  T result = Max(arg...);
  return Max(val, result);
}
template <typename T>
inline T read() {
  T X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  return w ? -X : X;
}
const int MAXN = 500005;
int n, m, k;
int deg[MAXN];
vector<int> G[MAXN];
int dep[MAXN];
int ans;
void dfs(int u, int p) {
  int cnt = 0;
  for (auto v : G[u]) {
    if (v == p) continue;
    dep[v] = dep[u] + 1;
    dfs(v, u);
    if (deg[v] == 1) cnt++;
  }
  ans -= max(0, cnt - 1);
}
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v), G[v].push_back(u);
    deg[u]++, deg[v]++;
  }
  int rt = 1;
  for (int i = 1; i <= n; i++) {
    if (deg[i] != 1) {
      rt = i;
      break;
    }
  }
  ans = n - 1;
  dfs(rt, rt);
  int odd = -1;
  bool flag = false;
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 1) {
      if (odd < 0)
        odd = dep[i] & 1;
      else if ((dep[i] & 1) != odd) {
        flag = true;
      }
    }
  }
  if (flag) {
    printf("3 %d\n", ans);
  } else {
    printf("1 %d\n", ans);
  }
}
int main() {
  solve();
  return 0;
}

```