#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using db = double;
using vi = vector<int>;
const int inf = 0x3f3f3f3f;
const db eps = 1e-8;
const int mod = 1e9 + 7;
ll qpow(ll a, ll b) {
  ll ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
const int maxn = 1e5 + 7;
int deg[maxn];
int n;
vector<int> G[maxn];
int ans1, ans2;
bool leaf(int u) { return deg[u] == 1; }
int dep[maxn];
void dfs(int u, int fa, int d) {
  dep[u] = d;
  int t = 0;
  for (auto v : G[u]) {
    if (v == fa) continue;
    dfs(v, u, d + 1);
    if (leaf(v)) t++;
  }
  if (t > 1) ans2 -= (t - 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1, u, v; i < n; ++i) {
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
    deg[u]++;
    deg[v]++;
  }
  vi a;
  for (int i = 1; i <= n; ++i) {
    if (deg[i] == 1) a.push_back(i);
  }
  ll num1 = 0, num2 = 0;
  ans2 = n - 1;
  dfs(1, 0, 1);
  for (auto u : a) {
    if (deg[u] == 1 && dep[u] & 1) num1++;
    if (deg[u] == 1 && dep[u] % 2 == 0) num2++;
  }
  if (num1 >= 1 && num2 > 0)
    ans1 = 3;
  else
    ans1 = 1;
  if (deg[1] == 1) {
    bool ok = 0;
    for (auto v : G[G[1][0]]) {
      if (v == 1) continue;
      if (leaf(v)) ok = 1;
    }
    if (ok) ans2--;
  }
  cout << ans1 << " " << ans2 << "\n";
}

### Prompt

Please create a solution in CPP 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;
using ll = long long;
using pii = pair<int, int>;
using db = double;
using vi = vector<int>;
const int inf = 0x3f3f3f3f;
const db eps = 1e-8;
const int mod = 1e9 + 7;
ll qpow(ll a, ll b) {
  ll ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
const int maxn = 1e5 + 7;
int deg[maxn];
int n;
vector<int> G[maxn];
int ans1, ans2;
bool leaf(int u) { return deg[u] == 1; }
int dep[maxn];
void dfs(int u, int fa, int d) {
  dep[u] = d;
  int t = 0;
  for (auto v : G[u]) {
    if (v == fa) continue;
    dfs(v, u, d + 1);
    if (leaf(v)) t++;
  }
  if (t > 1) ans2 -= (t - 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1, u, v; i < n; ++i) {
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
    deg[u]++;
    deg[v]++;
  }
  vi a;
  for (int i = 1; i <= n; ++i) {
    if (deg[i] == 1) a.push_back(i);
  }
  ll num1 = 0, num2 = 0;
  ans2 = n - 1;
  dfs(1, 0, 1);
  for (auto u : a) {
    if (deg[u] == 1 && dep[u] & 1) num1++;
    if (deg[u] == 1 && dep[u] % 2 == 0) num2++;
  }
  if (num1 >= 1 && num2 > 0)
    ans1 = 3;
  else
    ans1 = 1;
  if (deg[1] == 1) {
    bool ok = 0;
    for (auto v : G[G[1][0]]) {
      if (v == 1) continue;
      if (leaf(v)) ok = 1;
    }
    if (ok) ans2--;
  }
  cout << ans1 << " " << ans2 << "\n";
}

```