#include <bits/stdc++.h>
using namespace std;
typedef struct {
  int x;
  int y;
} H;
long long i, j, k, m, n, s, sz, t, x, y, p[100005], pv[100005], vi[100005],
    w[100005], z[100005];
H e[200040];
int C(H a, H b) { return a.x < b.x || a.x == b.x && a.y < b.y; }
void dfs(int v) {
  vi[v] = 1;
  for (int i = pv[v]; i < pv[v + 1]; i++) {
    if (!vi[e[i].y]) {
      p[e[i].y] = !p[v];
      dfs(e[i].y);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (; i < n - 1; i++) {
    cin >> e[i].x >> e[i].y;
    e[i + n - 1].x = e[i].y;
    e[i + n - 1].y = e[i].x;
    w[e[i].x]++;
    w[e[i].y]++;
  }
  sz = 2 * n - 2;
  sort(e, e + sz, C);
  for (i = 0; ++i < sz;) {
    if (e[i].x > e[i - 1].x) {
      for (j = e[i - 1].x; j++ < e[i].x;) {
        pv[j] = i;
      }
    }
  }
  for (j = e[sz - 1].x; j++ < n + 1;) pv[j] = sz;
  for (i = 0; i++ < n;) {
    if (w[i] < 2) {
      z[e[pv[i]].y]++;
    }
  }
  for (i = 0, y = n - 1; i++ < n;) {
    if (z[i] > 1) y -= z[i] - 1;
  }
  dfs(1);
  for (i = 0; i++ < n;) {
    if (z[i]) {
      if (k) {
        if (p[i] ^ p[k]) s = 1;
      } else {
        k = i;
      }
    }
  }
  if (s && n > 2)
    x = 3;
  else
    x = 1;
  cout << x << ' ' << y;
}

### 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;
typedef struct {
  int x;
  int y;
} H;
long long i, j, k, m, n, s, sz, t, x, y, p[100005], pv[100005], vi[100005],
    w[100005], z[100005];
H e[200040];
int C(H a, H b) { return a.x < b.x || a.x == b.x && a.y < b.y; }
void dfs(int v) {
  vi[v] = 1;
  for (int i = pv[v]; i < pv[v + 1]; i++) {
    if (!vi[e[i].y]) {
      p[e[i].y] = !p[v];
      dfs(e[i].y);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (; i < n - 1; i++) {
    cin >> e[i].x >> e[i].y;
    e[i + n - 1].x = e[i].y;
    e[i + n - 1].y = e[i].x;
    w[e[i].x]++;
    w[e[i].y]++;
  }
  sz = 2 * n - 2;
  sort(e, e + sz, C);
  for (i = 0; ++i < sz;) {
    if (e[i].x > e[i - 1].x) {
      for (j = e[i - 1].x; j++ < e[i].x;) {
        pv[j] = i;
      }
    }
  }
  for (j = e[sz - 1].x; j++ < n + 1;) pv[j] = sz;
  for (i = 0; i++ < n;) {
    if (w[i] < 2) {
      z[e[pv[i]].y]++;
    }
  }
  for (i = 0, y = n - 1; i++ < n;) {
    if (z[i] > 1) y -= z[i] - 1;
  }
  dfs(1);
  for (i = 0; i++ < n;) {
    if (z[i]) {
      if (k) {
        if (p[i] ^ p[k]) s = 1;
      } else {
        k = i;
      }
    }
  }
  if (s && n > 2)
    x = 3;
  else
    x = 1;
  cout << x << ' ' << y;
}

```