#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200000;
const long long INF = 1000000000ll;
const long long LLINF = 1000000000000000000ll;
const int MOD = 998244353;
vector<int> a[MAXN];
int deg[MAXN];
int dfsEq(int v, int pr) {
  int eq1 = -2;
  for (int u : a[v]) {
    if (u == pr) {
      continue;
    }
    int eq2 = dfsEq(u, v);
    if (eq1 == -2 || eq1 == eq2) {
      eq1 = eq2;
    } else {
      eq1 = -1;
    }
  }
  if (eq1 == -2) {
    eq1 = 0;
  } else {
    if (eq1 != -1) {
      eq1 = 1 - eq1;
    }
  }
  return eq1;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  int n;
  cin >> n;
  for (int i = 0; i < (int)(n - 1); i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    a[x].push_back(y);
    a[y].push_back(x);
    deg[x]++;
    deg[y]++;
  }
  int mn, mx;
  for (int i = 0; i < (int)(n); i++) {
    if (deg[i] > 1) {
      if (dfsEq(i, -1) != -1) {
        mn = 1;
      } else {
        mn = 3;
      }
      break;
    }
  }
  int cnt = 0;
  for (int v = 0; v < (int)(n); v++) {
    int c = 0;
    for (int u : a[v]) {
      if (deg[u] == 1) {
        c++;
      }
    }
    if (c > 0) {
      cnt += c - 1;
    }
  }
  mx = n - 1 - cnt;
  cout << mn << " " << mx << "\n";
  return 0;
}

### Prompt

Your challenge is to write 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;
const int MAXN = 200000;
const long long INF = 1000000000ll;
const long long LLINF = 1000000000000000000ll;
const int MOD = 998244353;
vector<int> a[MAXN];
int deg[MAXN];
int dfsEq(int v, int pr) {
  int eq1 = -2;
  for (int u : a[v]) {
    if (u == pr) {
      continue;
    }
    int eq2 = dfsEq(u, v);
    if (eq1 == -2 || eq1 == eq2) {
      eq1 = eq2;
    } else {
      eq1 = -1;
    }
  }
  if (eq1 == -2) {
    eq1 = 0;
  } else {
    if (eq1 != -1) {
      eq1 = 1 - eq1;
    }
  }
  return eq1;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  int n;
  cin >> n;
  for (int i = 0; i < (int)(n - 1); i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    a[x].push_back(y);
    a[y].push_back(x);
    deg[x]++;
    deg[y]++;
  }
  int mn, mx;
  for (int i = 0; i < (int)(n); i++) {
    if (deg[i] > 1) {
      if (dfsEq(i, -1) != -1) {
        mn = 1;
      } else {
        mn = 3;
      }
      break;
    }
  }
  int cnt = 0;
  for (int v = 0; v < (int)(n); v++) {
    int c = 0;
    for (int u : a[v]) {
      if (deg[u] == 1) {
        c++;
      }
    }
    if (c > 0) {
      cnt += c - 1;
    }
  }
  mx = n - 1 - cnt;
  cout << mn << " " << mx << "\n";
  return 0;
}

```