#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using db = double;
using ld = long double;
template <typename T>
using V = vector<T>;
template <typename T>
using VV = vector<vector<T>>;
constexpr ll mod = 1000000007;
constexpr ll INF = 1000000099;
constexpr ll LINF = (ll)(1e18 + 99);
const vector<ll> dx = {-1, 1, 0, 0}, dy = {0, 0, -1, 1};
template <typename T, typename U>
inline bool chmin(T& t, const U& u) {
  if (t > u) {
    t = u;
    return 1;
  }
  return 0;
}
template <typename T, typename U>
inline bool chmax(T& t, const U& u) {
  if (t < u) {
    t = u;
    return 1;
  }
  return 0;
}
template <typename T>
inline T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T, typename Y>
inline T mpow(T a, Y n) {
  T res = 1;
  for (; n; n >>= 1) {
    if (n & 1) res = res * a;
    a = a * a;
  }
  return res;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  ll n, lcnt = 0;
  cin >> n;
  VV<int> e(n, V<int>(0));
  set<int> lgr;
  bool ev = false, od = false;
  for (ll i = 0; i < (ll)(n - 1); ++i) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    e[a].emplace_back(b);
    e[b].emplace_back(a);
  }
  queue<int> q;
  q.push(0);
  q.push(-1);
  q.push(0);
  if ((ll)(e[0]).size() == 1) {
    lgr.insert(e[0][0]);
    ev = true;
    ++lcnt;
  }
  int now, dis, par;
  while (q.size()) {
    now = q.front();
    q.pop();
    par = q.front();
    q.pop();
    dis = q.front();
    q.pop();
    if (now && (ll)(e[now]).size() == 1) {
      lgr.insert(par);
      if (dis % 2)
        od = true;
      else
        ev = true;
      ++lcnt;
      cerr << now << '\n';
    } else {
      for (ll i = 0; i < (ll)((ll)(e[now]).size()); ++i) {
        if (e[now][i] == par) continue;
        q.push(e[now][i]);
        q.push(now);
        q.push(dis + 1);
      }
    }
  }
  if (!od || !ev)
    cout << 1 << " ";
  else
    cout << 3 << " ";
  cout << n - 1 - lcnt + (ll)(lgr).size() << '\n';
}

### 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>
using namespace std;
using ll = long long;
using db = double;
using ld = long double;
template <typename T>
using V = vector<T>;
template <typename T>
using VV = vector<vector<T>>;
constexpr ll mod = 1000000007;
constexpr ll INF = 1000000099;
constexpr ll LINF = (ll)(1e18 + 99);
const vector<ll> dx = {-1, 1, 0, 0}, dy = {0, 0, -1, 1};
template <typename T, typename U>
inline bool chmin(T& t, const U& u) {
  if (t > u) {
    t = u;
    return 1;
  }
  return 0;
}
template <typename T, typename U>
inline bool chmax(T& t, const U& u) {
  if (t < u) {
    t = u;
    return 1;
  }
  return 0;
}
template <typename T>
inline T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T, typename Y>
inline T mpow(T a, Y n) {
  T res = 1;
  for (; n; n >>= 1) {
    if (n & 1) res = res * a;
    a = a * a;
  }
  return res;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  ll n, lcnt = 0;
  cin >> n;
  VV<int> e(n, V<int>(0));
  set<int> lgr;
  bool ev = false, od = false;
  for (ll i = 0; i < (ll)(n - 1); ++i) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    e[a].emplace_back(b);
    e[b].emplace_back(a);
  }
  queue<int> q;
  q.push(0);
  q.push(-1);
  q.push(0);
  if ((ll)(e[0]).size() == 1) {
    lgr.insert(e[0][0]);
    ev = true;
    ++lcnt;
  }
  int now, dis, par;
  while (q.size()) {
    now = q.front();
    q.pop();
    par = q.front();
    q.pop();
    dis = q.front();
    q.pop();
    if (now && (ll)(e[now]).size() == 1) {
      lgr.insert(par);
      if (dis % 2)
        od = true;
      else
        ev = true;
      ++lcnt;
      cerr << now << '\n';
    } else {
      for (ll i = 0; i < (ll)((ll)(e[now]).size()); ++i) {
        if (e[now][i] == par) continue;
        q.push(e[now][i]);
        q.push(now);
        q.push(dis + 1);
      }
    }
  }
  if (!od || !ev)
    cout << 1 << " ";
  else
    cout << 3 << " ";
  cout << n - 1 - lcnt + (ll)(lgr).size() << '\n';
}

```