#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + c - '0';
  x *= f;
}
int n, rt;
int head[N], cnt;
int dep[N], f[N], du[N];
bool isleaf[N], visodd, viseven;
struct Edge {
  int to, next;
} e[N << 1];
void add(int x, int y) { e[++cnt] = (Edge){y, head[x]}, head[x] = cnt; }
void dfs(int u, int fa) {
  dep[u] = dep[fa] + 1;
  int lfcnt = 0;
  for (int i = head[u]; ~i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa) continue;
    if (isleaf[v]) {
      lfcnt++, dep[v] = dep[u] + 1;
      continue;
    }
    dfs(v, u);
    f[u] += f[v] + 1;
  }
  if (lfcnt) f[u]++;
}
int main() {
  memset(head, -1, sizeof(head));
  read(n);
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    read(x), read(y);
    add(x, y), add(y, x);
    du[x]++, du[y]++;
  }
  for (int i = 1; i <= n; i++)
    if (du[i] > 1) {
      rt = i;
      break;
    }
  for (int i = 1; i <= n; i++)
    if (du[i] == 1) isleaf[i] = true;
  dfs(rt, 0);
  for (int i = 1; i <= n; i++)
    if (isleaf[i]) {
      if (dep[i] & 1)
        visodd = true;
      else
        viseven = true;
    }
  if (visodd && viseven)
    printf("3 ");
  else
    printf("1 ");
  printf("%d\n", f[rt]);
  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 N = 1e5 + 5;
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + c - '0';
  x *= f;
}
int n, rt;
int head[N], cnt;
int dep[N], f[N], du[N];
bool isleaf[N], visodd, viseven;
struct Edge {
  int to, next;
} e[N << 1];
void add(int x, int y) { e[++cnt] = (Edge){y, head[x]}, head[x] = cnt; }
void dfs(int u, int fa) {
  dep[u] = dep[fa] + 1;
  int lfcnt = 0;
  for (int i = head[u]; ~i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa) continue;
    if (isleaf[v]) {
      lfcnt++, dep[v] = dep[u] + 1;
      continue;
    }
    dfs(v, u);
    f[u] += f[v] + 1;
  }
  if (lfcnt) f[u]++;
}
int main() {
  memset(head, -1, sizeof(head));
  read(n);
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    read(x), read(y);
    add(x, y), add(y, x);
    du[x]++, du[y]++;
  }
  for (int i = 1; i <= n; i++)
    if (du[i] > 1) {
      rt = i;
      break;
    }
  for (int i = 1; i <= n; i++)
    if (du[i] == 1) isleaf[i] = true;
  dfs(rt, 0);
  for (int i = 1; i <= n; i++)
    if (isleaf[i]) {
      if (dep[i] & 1)
        visodd = true;
      else
        viseven = true;
    }
  if (visodd && viseven)
    printf("3 ");
  else
    printf("1 ");
  printf("%d\n", f[rt]);
  return 0;
}

```