#include <bits/stdc++.h>
using namespace std;
int n, head[100005], tot, d[100005], fa[100005], vis[100005], lf[100005], cnt;
struct node {
  int v, nxt;
} edg[100005 << 1];
inline void addedg(int u, int v) {
  edg[tot].v = v;
  edg[tot].nxt = head[u];
  head[u] = tot++;
}
inline void init() {
  memset(head, -1, sizeof(int) * (n + 1));
  tot = cnt = 0;
}
inline void dfs1(int u, int f) {
  int v;
  d[u] = 0;
  for (int i = head[u]; i != -1; i = edg[i].nxt) {
    v = edg[i].v;
    ++d[u];
    if (v == f) continue;
    dfs1(v, u);
  }
}
inline void dfs2(int u, int f, int dis) {
  int v;
  fa[u] = f;
  d[u] = dis;
  for (int i = head[u]; i != -1; i = edg[i].nxt) {
    v = edg[i].v;
    if (f == v) continue;
    dfs2(v, u, dis + 1);
  }
}
inline bool cmp(int u, int v) { return d[u] > d[v]; }
set<int> se;
int main() {
  scanf("%d", &n);
  init();
  int u, v;
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    addedg(u, v);
    addedg(v, u);
  }
  dfs1(1, 1);
  for (int i = 1; i <= n; ++i) {
    if (d[i] == 1) lf[cnt++] = i;
  }
  int root = lf[0];
  dfs2(root, root, 0);
  vis[root] = 1;
  int maxn = 0, minn = 1;
  for (int i = 1; i < cnt; ++i) {
    if (d[lf[i]] & 1) minn = 3;
  }
  fa[root] = edg[head[root]].v;
  maxn = n - 1 - cnt;
  for (int i = 0; i < cnt; ++i) se.insert(fa[lf[i]]);
  maxn += se.size();
  printf("%d %d\n", minn, maxn);
  return 0;
}

### 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;
int n, head[100005], tot, d[100005], fa[100005], vis[100005], lf[100005], cnt;
struct node {
  int v, nxt;
} edg[100005 << 1];
inline void addedg(int u, int v) {
  edg[tot].v = v;
  edg[tot].nxt = head[u];
  head[u] = tot++;
}
inline void init() {
  memset(head, -1, sizeof(int) * (n + 1));
  tot = cnt = 0;
}
inline void dfs1(int u, int f) {
  int v;
  d[u] = 0;
  for (int i = head[u]; i != -1; i = edg[i].nxt) {
    v = edg[i].v;
    ++d[u];
    if (v == f) continue;
    dfs1(v, u);
  }
}
inline void dfs2(int u, int f, int dis) {
  int v;
  fa[u] = f;
  d[u] = dis;
  for (int i = head[u]; i != -1; i = edg[i].nxt) {
    v = edg[i].v;
    if (f == v) continue;
    dfs2(v, u, dis + 1);
  }
}
inline bool cmp(int u, int v) { return d[u] > d[v]; }
set<int> se;
int main() {
  scanf("%d", &n);
  init();
  int u, v;
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    addedg(u, v);
    addedg(v, u);
  }
  dfs1(1, 1);
  for (int i = 1; i <= n; ++i) {
    if (d[i] == 1) lf[cnt++] = i;
  }
  int root = lf[0];
  dfs2(root, root, 0);
  vis[root] = 1;
  int maxn = 0, minn = 1;
  for (int i = 1; i < cnt; ++i) {
    if (d[lf[i]] & 1) minn = 3;
  }
  fa[root] = edg[head[root]].v;
  maxn = n - 1 - cnt;
  for (int i = 0; i < cnt; ++i) se.insert(fa[lf[i]]);
  maxn += se.size();
  printf("%d %d\n", minn, maxn);
  return 0;
}

```