#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const double Pi = acos(-1);
namespace {
template <typename T>
inline void read(T &x) {
  x = 0;
  T f = 1;
  char s = getchar();
  for (; !isdigit(s); s = getchar())
    if (s == '-') f = -1;
  for (; isdigit(s); s = getchar()) x = (x << 3) + (x << 1) + (s ^ 48);
  x *= f;
}
}  // namespace
const int N = 1e5 + 5;
struct node {
  int to, next;
} e[N << 1];
int head[N], cnt, du[N];
bool vis[N];
void add(int u, int v) {
  e[++cnt] = node{v, head[u]};
  head[u] = cnt;
  e[++cnt] = node{u, head[v]};
  head[v] = cnt;
}
int siz[N];
void dfs(int u, int fa) {
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa) continue;
    du[v] = du[u] + 1;
    dfs(v, u);
  }
}
int main() {
  int n, u, v;
  read(n);
  for (register int i = (1); i < (n); ++i) {
    scanf("%d%d", &u, &v);
    add(u, v);
    du[u]++;
    du[v]++;
  }
  for (register int i = (1); i <= (n); ++i)
    if (du[i] == 1) vis[i] = 1, siz[e[head[i]].to]++;
  for (register int i = (1); i <= (n); ++i)
    if (du[i] != 1) {
      dfs(i, 0);
      break;
    }
  int f1 = 0, f2 = 0;
  for (register int i = (1); i <= (n); ++i) {
    if (!vis[i]) continue;
    if (du[i] & 1) f1 = 1;
    if (du[i] > 1 && du[i] % 2 == 0) f2 = 1;
  }
  int ans1 = 1, ans2 = n - 1;
  for (register int i = (1); i <= (n); ++i)
    if (siz[i]) ans2 -= siz[i] - 1;
  if (f1 && f2) ans1 = 3;
  cout << ans1 << " " << ans2 << endl;
}

### Prompt

Your task is to create 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;
const int INF = 0x3f3f3f3f;
const double Pi = acos(-1);
namespace {
template <typename T>
inline void read(T &x) {
  x = 0;
  T f = 1;
  char s = getchar();
  for (; !isdigit(s); s = getchar())
    if (s == '-') f = -1;
  for (; isdigit(s); s = getchar()) x = (x << 3) + (x << 1) + (s ^ 48);
  x *= f;
}
}  // namespace
const int N = 1e5 + 5;
struct node {
  int to, next;
} e[N << 1];
int head[N], cnt, du[N];
bool vis[N];
void add(int u, int v) {
  e[++cnt] = node{v, head[u]};
  head[u] = cnt;
  e[++cnt] = node{u, head[v]};
  head[v] = cnt;
}
int siz[N];
void dfs(int u, int fa) {
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa) continue;
    du[v] = du[u] + 1;
    dfs(v, u);
  }
}
int main() {
  int n, u, v;
  read(n);
  for (register int i = (1); i < (n); ++i) {
    scanf("%d%d", &u, &v);
    add(u, v);
    du[u]++;
    du[v]++;
  }
  for (register int i = (1); i <= (n); ++i)
    if (du[i] == 1) vis[i] = 1, siz[e[head[i]].to]++;
  for (register int i = (1); i <= (n); ++i)
    if (du[i] != 1) {
      dfs(i, 0);
      break;
    }
  int f1 = 0, f2 = 0;
  for (register int i = (1); i <= (n); ++i) {
    if (!vis[i]) continue;
    if (du[i] & 1) f1 = 1;
    if (du[i] > 1 && du[i] % 2 == 0) f2 = 1;
  }
  int ans1 = 1, ans2 = n - 1;
  for (register int i = (1); i <= (n); ++i)
    if (siz[i]) ans2 -= siz[i] - 1;
  if (f1 && f2) ans1 = 3;
  cout << ans1 << " " << ans2 << endl;
}

```