#include <bits/stdc++.h>
using namespace std;
long long addm(long long a, long long b, long long m) {
  return (a % m + b % m) % m;
}
long long mulm(long long a, long long b, long long m) {
  return (a % m * b % m) % m;
}
template <typename T>
void pvec(vector<T>& v) {
  for (int i = (0); i <= ((int)v.size() - 2); i++) cout << v[i] << " ";
  cout << v.back() << endl;
}
template <typename T>
void parray(T arr[], int n) {
  for (int i = (0); i <= (n - 2); i++) cout << arr[i] << " ";
  cout << arr[n - 1] << endl;
}
template <typename T>
void plist(list<T>& l) {
  vector<T> v(l.begin(), l.end());
  pvec(v);
}
template <typename T>
void mvec(vector<T>& v, T value) {
  for (int i = (0); i <= ((int)v.size() - 1); i++) v[i] = value;
}
template <typename T>
void marray(T arr[], T value, int n) {
  for (int i = (0); i <= (n - 1); i++) arr[i] = value;
}
template <typename T>
T sumarray(T a[], int n) {
  T res = 0;
  for (int i = (0); i <= (n - 1); i++) res += a[i];
  return res;
};
template <typename T>
T sumvec(vector<T> v) {
  T res = 0;
  for (int i = (0); i <= ((int)v.size() - 1); i++) res += v[i];
  return res;
}
template <typename T>
bool compare_seq(T a, int sa, T b, int sb) {
  for (int i = (0); i <= (min(sa, sb) - 1); i++)
    if (a[i] != b[i]) return a[i] < b[i];
  return sa < sb;
}
struct lexical_string {
  bool operator()(string a, string b) {
    return compare_seq<string>(a, a.length(), b, b.length());
  }
};
template <typename T>
struct lexical_vec {
  bool operator()(vector<T> a, vector<T> b) {
    return compare_seq<vector<T> >(a, (int)a.size(), b, (int)b.size());
  }
};
template <typename T>
struct VectorHash {
  size_t operator()(const std::vector<T>& v) const {
    std::hash<int> hasher;
    size_t seed = 0;
    for (T i : v) {
      seed ^= hasher(i) + 0x9e3779b9 + (seed << 6) + (seed >> 2);
    }
    return seed;
  }
};
vector<int> BFS(vector<vector<int> >& G, int node) {
  vector<int> res;
  queue<int> q;
  q.push(node);
  while (!q.empty()) {
    int current = q.front();
    q.pop();
    res.push_back(current);
    for (int c : G[current]) q.push(c);
  }
  return res;
}
vector<int> find_depth(vector<vector<int> >& H, int level[], int root) {
  int n = H.size();
  vector<int> depth(n, -1);
  vector<int> nodes = BFS(H, root);
  for (int idx = (nodes.size() - 1); idx >= (0); idx--) {
    int j = nodes[idx];
    if (H[j].size() == 0)
      depth[j] = 0;
    else {
      for (int child : H[j]) depth[j] = max(depth[j], 1 + depth[child]);
    }
  }
  return depth;
}
vector<vector<int> > get_children(vector<vector<int> >& G, int root) {
  int n = G.size();
  bool mark[n];
  marray(mark, false, n);
  queue<int> q;
  q.push(root);
  vector<vector<int> > children(n, vector<int>());
  while (!q.empty()) {
    int current = q.front();
    q.pop();
    mark[current] = true;
    for (int el : G[current]) {
      if (!mark[el]) {
        q.push(el);
        children[current].push_back(el);
      }
    }
  }
  return children;
}
vector<int> maxlength(vector<vector<int> >& H, vector<int>& depth, int root) {
  int n = H.size();
  vector<int> md(n, -1);
  vector<int> nodes = BFS(H, root);
  for (int idx = (nodes.size() - 1); idx >= (0); idx--) {
    int j = nodes[idx];
    if (depth[j] == 0)
      md[j] = 0;
    else if (depth[j] == 1)
      md[j] = max(depth[j], md[H[j][0]]);
    else {
      vector<int> cl;
      for (int child : H[j]) {
        cl.push_back(depth[child]);
      }
      sort(cl.begin(), cl.end());
      cout << "j: " << j << endl;
      pvec(cl);
      int cc = cl.size();
      md[j] = cl[cc - 1] + cl[cc - 2] + 2;
      for (int child : H[j]) {
        md[j] = max(md[j], md[child]);
      }
    }
  }
  return md;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<vector<int> > G(n, vector<int>());
  for (int i = (0); i <= (n - 2); i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  int root;
  for (int i = (0); i <= (n - 1); i++) {
    if (G[i].size() > 1) {
      root = i;
      break;
    }
  }
  vector<vector<int> > children = get_children(G, root);
  int level[n];
  bool mark[n];
  marray(mark, false, n);
  level[root] = 0;
  queue<int> q;
  q.push(root);
  while (!q.empty()) {
    int current = q.front();
    q.pop();
    mark[current] = true;
    for (int el : children[current]) {
      if (!mark[el]) {
        q.push(el);
        level[el] = level[current] + 1;
      }
    }
  }
  int maxnum = n - 1;
  for (int i = (0); i <= (n - 1); i++) {
    int count = 0;
    if (children[i].size() > 1) {
      for (int j : children[i])
        if (children[j].size() == 0) count++;
    }
    if (count > 1) maxnum -= (count - 1);
  }
  bool odd = false, even = false;
  for (int i = (0); i <= (n - 1); i++) {
    if (level[i] % 2 == 0 && children[i].empty()) even = true;
    if (level[i] % 2 == 1 && children[i].empty()) odd = true;
  }
  if (even && odd)
    cout << 3 << " " << maxnum << endl;
  else
    cout << 1 << " " << maxnum << endl;
  return 0;
}

### Prompt

Please formulate 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;
long long addm(long long a, long long b, long long m) {
  return (a % m + b % m) % m;
}
long long mulm(long long a, long long b, long long m) {
  return (a % m * b % m) % m;
}
template <typename T>
void pvec(vector<T>& v) {
  for (int i = (0); i <= ((int)v.size() - 2); i++) cout << v[i] << " ";
  cout << v.back() << endl;
}
template <typename T>
void parray(T arr[], int n) {
  for (int i = (0); i <= (n - 2); i++) cout << arr[i] << " ";
  cout << arr[n - 1] << endl;
}
template <typename T>
void plist(list<T>& l) {
  vector<T> v(l.begin(), l.end());
  pvec(v);
}
template <typename T>
void mvec(vector<T>& v, T value) {
  for (int i = (0); i <= ((int)v.size() - 1); i++) v[i] = value;
}
template <typename T>
void marray(T arr[], T value, int n) {
  for (int i = (0); i <= (n - 1); i++) arr[i] = value;
}
template <typename T>
T sumarray(T a[], int n) {
  T res = 0;
  for (int i = (0); i <= (n - 1); i++) res += a[i];
  return res;
};
template <typename T>
T sumvec(vector<T> v) {
  T res = 0;
  for (int i = (0); i <= ((int)v.size() - 1); i++) res += v[i];
  return res;
}
template <typename T>
bool compare_seq(T a, int sa, T b, int sb) {
  for (int i = (0); i <= (min(sa, sb) - 1); i++)
    if (a[i] != b[i]) return a[i] < b[i];
  return sa < sb;
}
struct lexical_string {
  bool operator()(string a, string b) {
    return compare_seq<string>(a, a.length(), b, b.length());
  }
};
template <typename T>
struct lexical_vec {
  bool operator()(vector<T> a, vector<T> b) {
    return compare_seq<vector<T> >(a, (int)a.size(), b, (int)b.size());
  }
};
template <typename T>
struct VectorHash {
  size_t operator()(const std::vector<T>& v) const {
    std::hash<int> hasher;
    size_t seed = 0;
    for (T i : v) {
      seed ^= hasher(i) + 0x9e3779b9 + (seed << 6) + (seed >> 2);
    }
    return seed;
  }
};
vector<int> BFS(vector<vector<int> >& G, int node) {
  vector<int> res;
  queue<int> q;
  q.push(node);
  while (!q.empty()) {
    int current = q.front();
    q.pop();
    res.push_back(current);
    for (int c : G[current]) q.push(c);
  }
  return res;
}
vector<int> find_depth(vector<vector<int> >& H, int level[], int root) {
  int n = H.size();
  vector<int> depth(n, -1);
  vector<int> nodes = BFS(H, root);
  for (int idx = (nodes.size() - 1); idx >= (0); idx--) {
    int j = nodes[idx];
    if (H[j].size() == 0)
      depth[j] = 0;
    else {
      for (int child : H[j]) depth[j] = max(depth[j], 1 + depth[child]);
    }
  }
  return depth;
}
vector<vector<int> > get_children(vector<vector<int> >& G, int root) {
  int n = G.size();
  bool mark[n];
  marray(mark, false, n);
  queue<int> q;
  q.push(root);
  vector<vector<int> > children(n, vector<int>());
  while (!q.empty()) {
    int current = q.front();
    q.pop();
    mark[current] = true;
    for (int el : G[current]) {
      if (!mark[el]) {
        q.push(el);
        children[current].push_back(el);
      }
    }
  }
  return children;
}
vector<int> maxlength(vector<vector<int> >& H, vector<int>& depth, int root) {
  int n = H.size();
  vector<int> md(n, -1);
  vector<int> nodes = BFS(H, root);
  for (int idx = (nodes.size() - 1); idx >= (0); idx--) {
    int j = nodes[idx];
    if (depth[j] == 0)
      md[j] = 0;
    else if (depth[j] == 1)
      md[j] = max(depth[j], md[H[j][0]]);
    else {
      vector<int> cl;
      for (int child : H[j]) {
        cl.push_back(depth[child]);
      }
      sort(cl.begin(), cl.end());
      cout << "j: " << j << endl;
      pvec(cl);
      int cc = cl.size();
      md[j] = cl[cc - 1] + cl[cc - 2] + 2;
      for (int child : H[j]) {
        md[j] = max(md[j], md[child]);
      }
    }
  }
  return md;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<vector<int> > G(n, vector<int>());
  for (int i = (0); i <= (n - 2); i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  int root;
  for (int i = (0); i <= (n - 1); i++) {
    if (G[i].size() > 1) {
      root = i;
      break;
    }
  }
  vector<vector<int> > children = get_children(G, root);
  int level[n];
  bool mark[n];
  marray(mark, false, n);
  level[root] = 0;
  queue<int> q;
  q.push(root);
  while (!q.empty()) {
    int current = q.front();
    q.pop();
    mark[current] = true;
    for (int el : children[current]) {
      if (!mark[el]) {
        q.push(el);
        level[el] = level[current] + 1;
      }
    }
  }
  int maxnum = n - 1;
  for (int i = (0); i <= (n - 1); i++) {
    int count = 0;
    if (children[i].size() > 1) {
      for (int j : children[i])
        if (children[j].size() == 0) count++;
    }
    if (count > 1) maxnum -= (count - 1);
  }
  bool odd = false, even = false;
  for (int i = (0); i <= (n - 1); i++) {
    if (level[i] % 2 == 0 && children[i].empty()) even = true;
    if (level[i] % 2 == 1 && children[i].empty()) odd = true;
  }
  if (even && odd)
    cout << 3 << " " << maxnum << endl;
  else
    cout << 1 << " " << maxnum << endl;
  return 0;
}

```