#include <bits/stdc++.h>
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
using namespace std;
const int N = 1e5 + 5;
int n, t, u, v;
vector<int> vec[N];
int num[N];
int juli[N];
int dfs(int rt, int fa, int woc) {
  if (vec[rt].size() > 2 || vec[rt].size() == 1) {
    return woc;
  }
  for (auto x : vec[rt]) {
    if (x != fa) {
      dfs(x, rt, woc + 1);
    }
  }
}
void dfs2(int rt, int fa, int shumu) {
  juli[rt] = shumu;
  for (auto x : vec[rt]) {
    if (x != fa) {
      dfs2(x, rt, shumu + 1);
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    vec[u].push_back(v);
    vec[v].push_back(u);
    num[u]++;
    num[v]++;
  }
  int maxx = n - 1;
  for (int i = 1; i <= n; i++) {
    int temp = 0;
    for (auto x : vec[i]) {
      if (num[x] == 1) {
        temp++;
      }
    }
    if (temp >= 2) {
      maxx -= (temp - 1);
    }
  }
  int ooooo = 0;
  int cnt = 0;
  int flag = 1;
  int pos = 0;
  for (int i = 1; i <= n; i++) {
    if (num[i] == 1) {
      pos = i;
      cnt++;
      if (cnt == 2) {
        if (dfs(i, 0, 1) % 2 != ooooo % 2) {
          flag = 0;
          break;
        }
      } else {
        ooooo += dfs(i, 0, 1);
      }
    }
  }
  if (flag == 0) {
    printf("3 ");
  } else {
    dfs2(pos, 0, 0);
    int nm = 0;
    for (int i = 1; i <= n; i++) {
      if (num[i] == 1 && juli[i] % 2 == 1) {
        nm = 1;
        break;
      }
    }
    if (nm)
      printf("3 ");
    else
      printf("1 ");
  }
  printf("%d\n", maxx);
  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>
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
using namespace std;
const int N = 1e5 + 5;
int n, t, u, v;
vector<int> vec[N];
int num[N];
int juli[N];
int dfs(int rt, int fa, int woc) {
  if (vec[rt].size() > 2 || vec[rt].size() == 1) {
    return woc;
  }
  for (auto x : vec[rt]) {
    if (x != fa) {
      dfs(x, rt, woc + 1);
    }
  }
}
void dfs2(int rt, int fa, int shumu) {
  juli[rt] = shumu;
  for (auto x : vec[rt]) {
    if (x != fa) {
      dfs2(x, rt, shumu + 1);
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    vec[u].push_back(v);
    vec[v].push_back(u);
    num[u]++;
    num[v]++;
  }
  int maxx = n - 1;
  for (int i = 1; i <= n; i++) {
    int temp = 0;
    for (auto x : vec[i]) {
      if (num[x] == 1) {
        temp++;
      }
    }
    if (temp >= 2) {
      maxx -= (temp - 1);
    }
  }
  int ooooo = 0;
  int cnt = 0;
  int flag = 1;
  int pos = 0;
  for (int i = 1; i <= n; i++) {
    if (num[i] == 1) {
      pos = i;
      cnt++;
      if (cnt == 2) {
        if (dfs(i, 0, 1) % 2 != ooooo % 2) {
          flag = 0;
          break;
        }
      } else {
        ooooo += dfs(i, 0, 1);
      }
    }
  }
  if (flag == 0) {
    printf("3 ");
  } else {
    dfs2(pos, 0, 0);
    int nm = 0;
    for (int i = 1; i <= n; i++) {
      if (num[i] == 1 && juli[i] % 2 == 1) {
        nm = 1;
        break;
      }
    }
    if (nm)
      printf("3 ");
    else
      printf("1 ");
  }
  printf("%d\n", maxx);
  return 0;
}

```