#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 10;
const long long mod = 2e18 + 7;
const int maxn = 1e5 + 7;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + (c - '0');
    c = getchar();
  }
  return x * f;
}
int head[maxn << 1];
struct Edge {
  int to, next;
} edge[maxn << 1];
int tot;
void init() {
  memset(head, -1, sizeof(head));
  tot = 0;
}
void addedge(int u, int v) {
  edge[tot].to = v;
  edge[tot].next = head[u];
  head[u] = tot++;
}
long long qpow(long long a, long long b, long long p) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % p;
    a = (a * a) % p;
    b >>= 1;
  }
  return res;
}
int deg[maxn];
int imin, imax;
void dfs(int u, int fa, int len) {
  if (deg[u] == 1) {
    if (len) imin = 3;
  }
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int v = edge[i].to;
    if (v == fa) continue;
    dfs(v, u, len ^ 1);
  }
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    init();
    memset(deg, 0, sizeof(deg));
    for (int i = (1); i <= (n - 1); i++) {
      int u, v;
      scanf("%d", &u), scanf("%d", &v);
      addedge(u, v), addedge(v, u);
      deg[u]++, deg[v]++;
    }
    imin = 1;
    for (int i = (1); i <= (n); i++) {
      if (deg[i] == 1) {
        dfs(i, 0, 0);
        break;
      }
    }
    imax = n - 1;
    for (int u = (1); u <= (n); u++) {
      int res = 0;
      for (int i = head[u]; i != -1; i = edge[i].next) {
        int v = edge[i].to;
        if (deg[v] == 1) {
          res++;
        }
      }
      if (res) imax -= res - 1;
    }
    printf("%d %d\n", imin, imax);
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
const int INF = 1e9 + 10;
const long long mod = 2e18 + 7;
const int maxn = 1e5 + 7;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + (c - '0');
    c = getchar();
  }
  return x * f;
}
int head[maxn << 1];
struct Edge {
  int to, next;
} edge[maxn << 1];
int tot;
void init() {
  memset(head, -1, sizeof(head));
  tot = 0;
}
void addedge(int u, int v) {
  edge[tot].to = v;
  edge[tot].next = head[u];
  head[u] = tot++;
}
long long qpow(long long a, long long b, long long p) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % p;
    a = (a * a) % p;
    b >>= 1;
  }
  return res;
}
int deg[maxn];
int imin, imax;
void dfs(int u, int fa, int len) {
  if (deg[u] == 1) {
    if (len) imin = 3;
  }
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int v = edge[i].to;
    if (v == fa) continue;
    dfs(v, u, len ^ 1);
  }
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    init();
    memset(deg, 0, sizeof(deg));
    for (int i = (1); i <= (n - 1); i++) {
      int u, v;
      scanf("%d", &u), scanf("%d", &v);
      addedge(u, v), addedge(v, u);
      deg[u]++, deg[v]++;
    }
    imin = 1;
    for (int i = (1); i <= (n); i++) {
      if (deg[i] == 1) {
        dfs(i, 0, 0);
        break;
      }
    }
    imax = n - 1;
    for (int u = (1); u <= (n); u++) {
      int res = 0;
      for (int i = head[u]; i != -1; i = edge[i].next) {
        int v = edge[i].to;
        if (deg[v] == 1) {
          res++;
        }
      }
      if (res) imax -= res - 1;
    }
    printf("%d %d\n", imin, imax);
  }
  return 0;
}

```