#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int out = 0;
  bool flag = false;
  register char cc = getchar();
  while (cc < '0' || cc > '9') {
    if (cc == '-') flag = true;
    cc = getchar();
  }
  while (cc >= '0' && cc <= '9') {
    out = (out << 3) + (out << 1) + (cc ^ 48);
    cc = getchar();
  }
  return flag ? -out : out;
}
inline void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x == 0)
    putchar('0');
  else {
    int num = 0;
    char cc[15];
    while (x) cc[++num] = x % 10 + 48, x /= 10;
    while (num) putchar(cc[num--]);
  }
  putchar('\n');
}
int N, d[100010], rt, head[100010], nxt[200010], to[200010], cnt, dep[100010],
    Min, Max;
bool a, b;
void add(int u, int v) {
  to[++cnt] = v;
  nxt[cnt] = head[u];
  head[u] = cnt;
}
void dfs(int u, int fa) {
  dep[u] = dep[fa] + 1;
  if (d[u] != 1)
    Max++;
  else if (dep[u] & 1)
    a = true;
  else if (!(dep[u] & 1))
    b = true;
  bool o = false;
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (v == fa) continue;
    if (d[v] == 1) o = true;
    dfs(v, u);
  }
  if (o) Max++;
}
int main() {
  N = read();
  for (int i = 1; i < N; i++) {
    int u = read(), v = read();
    d[u]++, d[v]++;
    add(u, v), add(v, u);
  }
  for (int i = 1; i <= N; i++)
    if (d[i] != 1) {
      rt = i;
      break;
    }
  dfs(rt, 0);
  if (a && b)
    Min = 3;
  else
    Min = 1;
  cout << Min << ' ' << Max - 1 << endl;
  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;
inline int read() {
  int out = 0;
  bool flag = false;
  register char cc = getchar();
  while (cc < '0' || cc > '9') {
    if (cc == '-') flag = true;
    cc = getchar();
  }
  while (cc >= '0' && cc <= '9') {
    out = (out << 3) + (out << 1) + (cc ^ 48);
    cc = getchar();
  }
  return flag ? -out : out;
}
inline void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x == 0)
    putchar('0');
  else {
    int num = 0;
    char cc[15];
    while (x) cc[++num] = x % 10 + 48, x /= 10;
    while (num) putchar(cc[num--]);
  }
  putchar('\n');
}
int N, d[100010], rt, head[100010], nxt[200010], to[200010], cnt, dep[100010],
    Min, Max;
bool a, b;
void add(int u, int v) {
  to[++cnt] = v;
  nxt[cnt] = head[u];
  head[u] = cnt;
}
void dfs(int u, int fa) {
  dep[u] = dep[fa] + 1;
  if (d[u] != 1)
    Max++;
  else if (dep[u] & 1)
    a = true;
  else if (!(dep[u] & 1))
    b = true;
  bool o = false;
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (v == fa) continue;
    if (d[v] == 1) o = true;
    dfs(v, u);
  }
  if (o) Max++;
}
int main() {
  N = read();
  for (int i = 1; i < N; i++) {
    int u = read(), v = read();
    d[u]++, d[v]++;
    add(u, v), add(v, u);
  }
  for (int i = 1; i <= N; i++)
    if (d[i] != 1) {
      rt = i;
      break;
    }
  dfs(rt, 0);
  if (a && b)
    Min = 3;
  else
    Min = 1;
  cout << Min << ' ' << Max - 1 << endl;
  return 0;
}

```