#include <bits/stdc++.h>
using namespace std;
int n, tot;
int eg[100005][2];
int deg[100005], mark[100005], ans[100005];
struct edge {
  int v;
  edge *next;
} pool[200005], *h[100005];
void addedge(int u, int v) {
  edge *p = &pool[tot++];
  p->v = v;
  p->next = h[u];
  h[u] = p;
}
void dfs(int u, int fa) {
  ans[u] = 1 - ans[fa];
  for (edge *p = h[u]; p; p = p->next) {
    if (p->v != fa) dfs(p->v, u);
  }
}
bool chk() {
  int tmp = -1;
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 1) {
      if (tmp == -1)
        tmp = ans[i];
      else if (ans[i] != tmp)
        return 0;
    }
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &eg[i][0], &eg[i][1]);
    deg[eg[i][0]]++, deg[eg[i][1]]++;
    addedge(eg[i][0], eg[i][1]);
    addedge(eg[i][1], eg[i][0]);
  }
  dfs(1, 0);
  for (int i = 1; i < n; i++) {
    if (deg[eg[i][0]] == 1) mark[eg[i][1]] = 1;
    if (deg[eg[i][1]] == 1) mark[eg[i][0]] = 1;
  }
  int cnt = n - 1;
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 1) cnt--;
    if (mark[i] == 1) cnt++;
  }
  if (chk())
    printf("1 %d", cnt);
  else
    printf("3 %d", cnt);
  return 0;
}

### Prompt

Generate 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, tot;
int eg[100005][2];
int deg[100005], mark[100005], ans[100005];
struct edge {
  int v;
  edge *next;
} pool[200005], *h[100005];
void addedge(int u, int v) {
  edge *p = &pool[tot++];
  p->v = v;
  p->next = h[u];
  h[u] = p;
}
void dfs(int u, int fa) {
  ans[u] = 1 - ans[fa];
  for (edge *p = h[u]; p; p = p->next) {
    if (p->v != fa) dfs(p->v, u);
  }
}
bool chk() {
  int tmp = -1;
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 1) {
      if (tmp == -1)
        tmp = ans[i];
      else if (ans[i] != tmp)
        return 0;
    }
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &eg[i][0], &eg[i][1]);
    deg[eg[i][0]]++, deg[eg[i][1]]++;
    addedge(eg[i][0], eg[i][1]);
    addedge(eg[i][1], eg[i][0]);
  }
  dfs(1, 0);
  for (int i = 1; i < n; i++) {
    if (deg[eg[i][0]] == 1) mark[eg[i][1]] = 1;
    if (deg[eg[i][1]] == 1) mark[eg[i][0]] = 1;
  }
  int cnt = n - 1;
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 1) cnt--;
    if (mark[i] == 1) cnt++;
  }
  if (chk())
    printf("1 %d", cnt);
  else
    printf("3 %d", cnt);
  return 0;
}

```