#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int n, cnt_edge, ans = 1;
int head[maxn], deg[maxn];
bool f[maxn][2];
struct edge {
  int to, nxt;
} e[maxn << 1];
inline int read() {
  char c = getchar();
  int res = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') res = res * 10 + c - '0', c = getchar();
  return res * f;
}
inline void add_edge(int u, int v) {
  e[++cnt_edge].nxt = head[u];
  head[u] = cnt_edge;
  e[cnt_edge].to = v;
}
void dfs1(int x, int fa) {
  if (!e[head[x]].nxt) f[x][0] = 1;
  for (int i = head[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    if (y == fa) continue;
    dfs1(y, x);
    if (f[x][0] && f[y][0]) ans = 3;
    if (f[x][1] && f[y][1]) ans = 3;
    f[x][0] |= f[y][1], f[x][1] |= f[y][0];
  }
}
void dfs2(int x, int fa) {
  if (!e[head[x]].nxt) {
    if (x == 1)
      deg[e[head[x]].to]++;
    else {
      deg[fa]++;
      return;
    }
  }
  for (int i = head[x]; i; i = e[i].nxt)
    if (e[i].to != fa) dfs2(e[i].to, x);
}
int main() {
  n = read();
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    add_edge(u, v);
    add_edge(v, u);
  }
  dfs1(1, 0);
  printf("%d ", ans);
  ans = n - 1;
  dfs2(1, 0);
  for (int i = 1; i <= n; i++)
    if (deg[i]) ans -= deg[i] - 1;
  printf("%d", ans);
  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 maxn = 1e5 + 10;
int n, cnt_edge, ans = 1;
int head[maxn], deg[maxn];
bool f[maxn][2];
struct edge {
  int to, nxt;
} e[maxn << 1];
inline int read() {
  char c = getchar();
  int res = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') res = res * 10 + c - '0', c = getchar();
  return res * f;
}
inline void add_edge(int u, int v) {
  e[++cnt_edge].nxt = head[u];
  head[u] = cnt_edge;
  e[cnt_edge].to = v;
}
void dfs1(int x, int fa) {
  if (!e[head[x]].nxt) f[x][0] = 1;
  for (int i = head[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    if (y == fa) continue;
    dfs1(y, x);
    if (f[x][0] && f[y][0]) ans = 3;
    if (f[x][1] && f[y][1]) ans = 3;
    f[x][0] |= f[y][1], f[x][1] |= f[y][0];
  }
}
void dfs2(int x, int fa) {
  if (!e[head[x]].nxt) {
    if (x == 1)
      deg[e[head[x]].to]++;
    else {
      deg[fa]++;
      return;
    }
  }
  for (int i = head[x]; i; i = e[i].nxt)
    if (e[i].to != fa) dfs2(e[i].to, x);
}
int main() {
  n = read();
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    add_edge(u, v);
    add_edge(v, u);
  }
  dfs1(1, 0);
  printf("%d ", ans);
  ans = n - 1;
  dfs2(1, 0);
  for (int i = 1; i <= n; i++)
    if (deg[i]) ans -= deg[i] - 1;
  printf("%d", ans);
  return 0;
}

```