#include <bits/stdc++.h>
using namespace std;
char p[30] = {'0', 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h',
              'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q',
              'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
inline int rd() {
  int x = 0, flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch ^ 48);
    ch = getchar();
  }
  return x * flag;
}
int wl, x, y, head[100010], mrk[100010], ans, dep[100010];
vector<int> lv;
struct node {
  int nxt, to;
} star[100010 << 2];
inline void add(int from, int to) {
  wl++;
  star[wl].to = to;
  star[wl].nxt = head[from];
  head[from] = wl;
}
int in[100010], rt;
void dfs(int now, int fa) {
  dep[now] = dep[fa] + 1;
  int ok = 0;
  for (register int i = head[now]; i; i = star[i].nxt) {
    int u = star[i].to;
    if (u == fa) continue;
    dfs(u, now);
    if (mrk[u] == 1) ans -= 1, ok = 1;
  }
  if (ok) ans += 1;
}
inline int lowbit(int x) { return x & (-x); }
int n;
int main() {
  n = rd();
  ans = n - 1;
  for (register int i = 1; i < n; ++i) {
    x = rd(), y = rd();
    add(x, y), add(y, x);
    in[x]++, in[y]++;
  }
  for (register int i = 1; i <= n; ++i) {
    if (in[i] >= 2) rt = i;
    if (in[i] == 1) lv.push_back(i), mrk[i] = 1;
  }
  dfs(rt, 0);
  int mn = 1;
  for (register int i = 1; i < lv.size(); ++i) {
    if ((dep[lv[i]] + dep[lv[i - 1]]) % 2 == 1) mn = 3;
  }
  cout << mn << " " << ans;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
char p[30] = {'0', 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h',
              'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q',
              'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
inline int rd() {
  int x = 0, flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch ^ 48);
    ch = getchar();
  }
  return x * flag;
}
int wl, x, y, head[100010], mrk[100010], ans, dep[100010];
vector<int> lv;
struct node {
  int nxt, to;
} star[100010 << 2];
inline void add(int from, int to) {
  wl++;
  star[wl].to = to;
  star[wl].nxt = head[from];
  head[from] = wl;
}
int in[100010], rt;
void dfs(int now, int fa) {
  dep[now] = dep[fa] + 1;
  int ok = 0;
  for (register int i = head[now]; i; i = star[i].nxt) {
    int u = star[i].to;
    if (u == fa) continue;
    dfs(u, now);
    if (mrk[u] == 1) ans -= 1, ok = 1;
  }
  if (ok) ans += 1;
}
inline int lowbit(int x) { return x & (-x); }
int n;
int main() {
  n = rd();
  ans = n - 1;
  for (register int i = 1; i < n; ++i) {
    x = rd(), y = rd();
    add(x, y), add(y, x);
    in[x]++, in[y]++;
  }
  for (register int i = 1; i <= n; ++i) {
    if (in[i] >= 2) rt = i;
    if (in[i] == 1) lv.push_back(i), mrk[i] = 1;
  }
  dfs(rt, 0);
  int mn = 1;
  for (register int i = 1; i < lv.size(); ++i) {
    if ((dep[lv[i]] + dep[lv[i - 1]]) % 2 == 1) mn = 3;
  }
  cout << mn << " " << ans;
  return 0;
}

```