#include <bits/stdc++.h>
using namespace std;
int a[200005], b[200005], g[200005], dgr[200005], tot, head[200005],
    dep[200005], pr[200005];
struct edge {
  int to, next;
} e[200005 << 1];
void add(int x, int y) {
  tot++;
  e[tot].next = head[x];
  e[tot].to = y;
  head[x] = tot;
}
void dfs(int x, int fr) {
  dep[x] = dep[fr] + 1;
  pr[x] = fr;
  for (int p = head[x]; p; p = e[p].next) {
    int v = e[p].to;
    if (v == fr) continue;
    dfs(v, x);
  }
}
void solve() {
  int n;
  scanf("%d", &n);
  tot = 0;
  memset(dgr, 0, n * sizeof(dgr[0]));
  memset(head + 1, 0, n * sizeof(head[0]));
  for (int i = 1; i < n; i++) {
    int f, g;
    scanf("%d %d", &f, &g);
    dgr[f]++;
    dgr[g]++;
    add(f, g);
    add(g, f);
  }
  if (n == 2) {
    printf("1 1\n");
    return;
  }
  int ans = n;
  bool ex = 0;
  for (int i = 1; i <= n; i++) {
    int cnt = 0;
    if (dgr[i] == 1) continue;
    bool nex = 0;
    for (int p = head[i]; p; p = e[p].next) {
      int v = e[p].to;
      if (dgr[v] == 1) {
        nex = 1;
        cnt++;
      }
    }
    if (nex = 0) ex = 1;
    if (!cnt) continue;
    ans -= cnt - 1;
  }
  ans--;
  memset(dep, 0, sizeof(dep));
  dep[1] = -1;
  dfs(1, 1);
  int mx = 0, mi = 0;
  for (int i = 1; i <= n; i++) {
    if (dep[i] > mx) {
      mx = dep[i];
      mi = i;
    }
  }
  dep[mi] = -1;
  dfs(mi, mi);
  int mii = mi;
  mx = 0;
  for (int i = 1; i <= n; i++) {
    if (dep[i] > mx) {
      mx = dep[i];
      mi = i;
    }
  }
  if (mx % 2) {
    printf("3 %d\n", ans);
    return;
  }
  int r = mi;
  for (int i = 1; i <= mx / 2; i++) {
    r = pr[r];
  }
  dep[r] = -1;
  dfs(r, r);
  int o = 0, e = 0, on = 0;
  for (int i = 1; i <= n; i++) {
    if (dgr[i] != 1) continue;
    if (dep[i] % 2 == 0)
      e++;
    else if (dep[i] == 1)
      on++;
    else
      o++;
  }
  if (!e || !(on + o))
    printf("%d %d\n", 1, ans);
  else {
    printf("%d %d\n", 3, ans);
  }
}
int main() {
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 a[200005], b[200005], g[200005], dgr[200005], tot, head[200005],
    dep[200005], pr[200005];
struct edge {
  int to, next;
} e[200005 << 1];
void add(int x, int y) {
  tot++;
  e[tot].next = head[x];
  e[tot].to = y;
  head[x] = tot;
}
void dfs(int x, int fr) {
  dep[x] = dep[fr] + 1;
  pr[x] = fr;
  for (int p = head[x]; p; p = e[p].next) {
    int v = e[p].to;
    if (v == fr) continue;
    dfs(v, x);
  }
}
void solve() {
  int n;
  scanf("%d", &n);
  tot = 0;
  memset(dgr, 0, n * sizeof(dgr[0]));
  memset(head + 1, 0, n * sizeof(head[0]));
  for (int i = 1; i < n; i++) {
    int f, g;
    scanf("%d %d", &f, &g);
    dgr[f]++;
    dgr[g]++;
    add(f, g);
    add(g, f);
  }
  if (n == 2) {
    printf("1 1\n");
    return;
  }
  int ans = n;
  bool ex = 0;
  for (int i = 1; i <= n; i++) {
    int cnt = 0;
    if (dgr[i] == 1) continue;
    bool nex = 0;
    for (int p = head[i]; p; p = e[p].next) {
      int v = e[p].to;
      if (dgr[v] == 1) {
        nex = 1;
        cnt++;
      }
    }
    if (nex = 0) ex = 1;
    if (!cnt) continue;
    ans -= cnt - 1;
  }
  ans--;
  memset(dep, 0, sizeof(dep));
  dep[1] = -1;
  dfs(1, 1);
  int mx = 0, mi = 0;
  for (int i = 1; i <= n; i++) {
    if (dep[i] > mx) {
      mx = dep[i];
      mi = i;
    }
  }
  dep[mi] = -1;
  dfs(mi, mi);
  int mii = mi;
  mx = 0;
  for (int i = 1; i <= n; i++) {
    if (dep[i] > mx) {
      mx = dep[i];
      mi = i;
    }
  }
  if (mx % 2) {
    printf("3 %d\n", ans);
    return;
  }
  int r = mi;
  for (int i = 1; i <= mx / 2; i++) {
    r = pr[r];
  }
  dep[r] = -1;
  dfs(r, r);
  int o = 0, e = 0, on = 0;
  for (int i = 1; i <= n; i++) {
    if (dgr[i] != 1) continue;
    if (dep[i] % 2 == 0)
      e++;
    else if (dep[i] == 1)
      on++;
    else
      o++;
  }
  if (!e || !(on + o))
    printf("%d %d\n", 1, ans);
  else {
    printf("%d %d\n", 3, ans);
  }
}
int main() {
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

```