#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, w = 1;
  char ch = 0;
  while (ch > '9' || ch < '0') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * w;
}
const int N = 3e5 + 5;
struct edge {
  int next, to;
} a[N << 1];
int head[N], cnt;
inline void add(int u, int v) {
  a[++cnt].to = v;
  a[cnt].next = head[u];
  head[u] = cnt;
}
int n, ans1, ans2;
int deg[N], dep[N], flag[2];
int f[N];
int find(int x) { return (x == f[x]) ? x : f[x] = find(f[x]); }
inline void link(int x, int y) {
  x = find(x), y = find(y);
  f[x] = y;
}
void dfs(int x, int fa) {
  dep[x] = dep[fa] + 1;
  if (deg[x] == 1) flag[dep[x] & 1] = 1;
  int las = 0;
  for (int i = head[x]; i; i = a[i].next) {
    int y = a[i].to;
    if (y == fa) continue;
    if (deg[y] == 1 && las) link(las, y);
    if (deg[y] == 1) las = y;
    dfs(y, x);
  }
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) f[i] = i;
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read();
    ++deg[u], ++deg[v];
    add(u, v), add(v, u);
  }
  for (int i = 1; i <= n; ++i)
    if (deg[i] != 1) {
      dfs(i, 0);
      for (int j = 1; j <= n; ++j)
        if (i != j && find(j) == j) ++ans2;
      break;
    }
  if (flag[1] & flag[0])
    ans1 = 3;
  else
    ans1 = 1;
  printf("%d %d\n", ans1, ans2);
  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;
inline int read() {
  int x = 0, w = 1;
  char ch = 0;
  while (ch > '9' || ch < '0') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * w;
}
const int N = 3e5 + 5;
struct edge {
  int next, to;
} a[N << 1];
int head[N], cnt;
inline void add(int u, int v) {
  a[++cnt].to = v;
  a[cnt].next = head[u];
  head[u] = cnt;
}
int n, ans1, ans2;
int deg[N], dep[N], flag[2];
int f[N];
int find(int x) { return (x == f[x]) ? x : f[x] = find(f[x]); }
inline void link(int x, int y) {
  x = find(x), y = find(y);
  f[x] = y;
}
void dfs(int x, int fa) {
  dep[x] = dep[fa] + 1;
  if (deg[x] == 1) flag[dep[x] & 1] = 1;
  int las = 0;
  for (int i = head[x]; i; i = a[i].next) {
    int y = a[i].to;
    if (y == fa) continue;
    if (deg[y] == 1 && las) link(las, y);
    if (deg[y] == 1) las = y;
    dfs(y, x);
  }
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) f[i] = i;
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read();
    ++deg[u], ++deg[v];
    add(u, v), add(v, u);
  }
  for (int i = 1; i <= n; ++i)
    if (deg[i] != 1) {
      dfs(i, 0);
      for (int j = 1; j <= n; ++j)
        if (i != j && find(j) == j) ++ans2;
      break;
    }
  if (flag[1] & flag[0])
    ans1 = 3;
  else
    ans1 = 1;
  printf("%d %d\n", ans1, ans2);
  return 0;
}

```