Problem: Berland is very concerned with privacy, so almost all plans and blueprints are secret. However, a spy of the neighboring state managed to steal the Bertown subway scheme.

The Bertown Subway has n stations, numbered from 1 to n, and m bidirectional tunnels connecting them. All Bertown Subway consists of lines. To be more precise, there are two types of lines: circular and radial.

A radial line is a sequence of stations v1, ..., vk (k > 1), where stations vi and vi + 1 (i < k) are connected by a tunnel and no station occurs in the line more than once (vi ≠ vj for i ≠ j).

A loop line is a series of stations, v1, ..., vk (k > 2), where stations vi и vi + 1 are connected by a tunnel. In addition, stations v1 and vk are also connected by a tunnel. No station is occurs in the loop line more than once.

Note that a single station can be passed by any number of lines.

According to Berland standards, there can't be more than one tunnel between two stations and each tunnel belongs to exactly one line. Naturally, each line has at least one tunnel. Between any two stations there is the way along the subway tunnels. In addition, in terms of graph theory, a subway is a vertex cactus: if we consider the subway as a graph in which the stations are the vertexes and the edges are tunnels, then each vertex lies on no more than one simple cycle.

Unfortunately, scheme, stolen by the spy, had only the stations and the tunnels. It was impossible to determine to which line every tunnel corresponds. But to sabotage successfully, the spy needs to know what minimum and maximum number of lines may be in the Bertown subway.

Help him!

Input

The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ 3·105) — the number of stations and the number of tunnels, correspondingly.

Each of the next m lines contain two integers — the numbers of stations connected by the corresponding tunnel. The stations are numbered with integers from 1 to n.

It is guaranteed that the graph that corresponds to the subway has no multiple edges or loops, it is connected and it is a vertex cactus.

Output

Print two numbers — the minimum and maximum number of lines correspondingly.

Examples

Input

3 3
1 2
2 3
3 1


Output

1 3


Input

8 8
1 2
2 3
3 4
4 5
6 4
4 7
7 2
2 8


Output

2 8


Input

6 6
1 2
2 3
2 5
5 6
3 4
3 5


Output

3 6

Note

The subway scheme with minimum possible number of lines for the second sample is: 

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:36777216")
template <class T>
inline void RD(T &);
template <class T>
inline void OT(const T &);
inline int RD() {
  int x;
  RD(x);
  return x;
}
template <class T>
inline T &_RD(T &x) {
  RD(x);
  return x;
}
inline void RC(char &c) { scanf(" %c", &c); }
inline void RS(char *s) { scanf("%s", s); }
template <class T0, class T1>
inline void RD(T0 &x0, T1 &x1) {
  RD(x0), RD(x1);
}
template <class T0, class T1, class T2>
inline void RD(T0 &x0, T1 &x1, T2 &x2) {
  RD(x0), RD(x1), RD(x2);
}
template <class T0, class T1, class T2, class T3>
inline void RD(T0 &x0, T1 &x1, T2 &x2, T3 &x3) {
  RD(x0), RD(x1), RD(x2), RD(x3);
}
template <class T0, class T1, class T2, class T3, class T4>
inline void RD(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4) {
  RD(x0), RD(x1), RD(x2), RD(x3), RD(x4);
}
template <class T0, class T1, class T2, class T3, class T4, class T5>
inline void RD(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4, T5 &x5) {
  RD(x0), RD(x1), RD(x2), RD(x3), RD(x4), RD(x5);
}
template <class T0, class T1, class T2, class T3, class T4, class T5, class T6>
inline void RD(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4, T5 &x5, T6 &x6) {
  RD(x0), RD(x1), RD(x2), RD(x3), RD(x4), RD(x5), RD(x6);
}
template <class T0, class T1>
inline void OT(T0 &x0, T1 &x1) {
  OT(x0), OT(x1);
}
template <class T0, class T1, class T2>
inline void OT(T0 &x0, T1 &x1, T2 &x2) {
  OT(x0), OT(x1), OT(x2);
}
template <class T0, class T1, class T2, class T3>
inline void OT(T0 &x0, T1 &x1, T2 &x2, T3 &x3) {
  OT(x0), OT(x1), OT(x2), OT(x3);
}
template <class T0, class T1, class T2, class T3, class T4>
inline void OT(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4) {
  OT(x0), OT(x1), OT(x2), OT(x3), OT(x4);
}
template <class T0, class T1, class T2, class T3, class T4, class T5>
inline void OT(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4, T5 &x5) {
  OT(x0), OT(x1), OT(x2), OT(x3), OT(x4), OT(x5);
}
template <class T0, class T1, class T2, class T3, class T4, class T5, class T6>
inline void OT(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4, T5 &x5, T6 &x6) {
  OT(x0), OT(x1), OT(x2), OT(x3), OT(x4), OT(x5), OT(x6);
}
template <class T>
inline void RST(T &A) {
  memset(A, 0, sizeof(A));
}
template <class T0, class T1>
inline void RST(T0 &A0, T1 &A1) {
  RST(A0), RST(A1);
}
template <class T0, class T1, class T2>
inline void RST(T0 &A0, T1 &A1, T2 &A2) {
  RST(A0), RST(A1), RST(A2);
}
template <class T0, class T1, class T2, class T3>
inline void RST(T0 &A0, T1 &A1, T2 &A2, T3 &A3) {
  RST(A0), RST(A1), RST(A2), RST(A3);
}
template <class T0, class T1, class T2, class T3, class T4>
inline void RST(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4) {
  RST(A0), RST(A1), RST(A2), RST(A3), RST(A4);
}
template <class T0, class T1, class T2, class T3, class T4, class T5>
inline void RST(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5) {
  RST(A0), RST(A1), RST(A2), RST(A3), RST(A4), RST(A5);
}
template <class T0, class T1, class T2, class T3, class T4, class T5, class T6>
inline void RST(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5, T6 &A6) {
  RST(A0), RST(A1), RST(A2), RST(A3), RST(A4), RST(A5), RST(A6);
}
template <class T>
inline void CLR(priority_queue<T, vector<T>, less<T> > &Q) {
  while (!Q.empty()) Q.pop();
}
template <class T>
inline void CLR(priority_queue<T, vector<T>, greater<T> > &Q) {
  while (!Q.empty()) Q.pop();
}
template <class T>
inline void CLR(T &A) {
  A.clear();
}
template <class T0, class T1>
inline void CLR(T0 &A0, T1 &A1) {
  CLR(A0), CLR(A1);
}
template <class T0, class T1, class T2>
inline void CLR(T0 &A0, T1 &A1, T2 &A2) {
  CLR(A0), CLR(A1), CLR(A2);
}
template <class T0, class T1, class T2, class T3>
inline void CLR(T0 &A0, T1 &A1, T2 &A2, T3 &A3) {
  CLR(A0), CLR(A1), CLR(A2), CLR(A3);
}
template <class T0, class T1, class T2, class T3, class T4>
inline void CLR(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4) {
  CLR(A0), CLR(A1), CLR(A2), CLR(A3), CLR(A4);
}
template <class T0, class T1, class T2, class T3, class T4, class T5>
inline void CLR(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5) {
  CLR(A0), CLR(A1), CLR(A2), CLR(A3), CLR(A4), CLR(A5);
}
template <class T0, class T1, class T2, class T3, class T4, class T5, class T6>
inline void CLR(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5, T6 &A6) {
  CLR(A0), CLR(A1), CLR(A2), CLR(A3), CLR(A4), CLR(A5), CLR(A6);
}
template <class T>
inline void CLR(T &A, int n) {
  for (int i = 0; i < int(n); ++i) CLR(A[i]);
}
template <class T>
inline void FLC(T &A, int x) {
  memset(A, x, sizeof(A));
}
template <class T0, class T1>
inline void FLC(T0 &A0, T1 &A1, int x) {
  FLC(A0, x), FLC(A1, x);
}
template <class T0, class T1, class T2>
inline void FLC(T0 &A0, T1 &A1, T2 &A2) {
  FLC(A0), FLC(A1), FLC(A2);
}
template <class T0, class T1, class T2, class T3>
inline void FLC(T0 &A0, T1 &A1, T2 &A2, T3 &A3) {
  FLC(A0), FLC(A1), FLC(A2), FLC(A3);
}
template <class T0, class T1, class T2, class T3, class T4>
inline void FLC(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4) {
  FLC(A0), FLC(A1), FLC(A2), FLC(A3), FLC(A4);
}
template <class T0, class T1, class T2, class T3, class T4, class T5>
inline void FLC(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5) {
  FLC(A0), FLC(A1), FLC(A2), FLC(A3), FLC(A4), FLC(A5);
}
template <class T0, class T1, class T2, class T3, class T4, class T5, class T6>
inline void FLC(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5, T6 &A6) {
  FLC(A0), FLC(A1), FLC(A2), FLC(A3), FLC(A4), FLC(A5), FLC(A6);
}
template <class T>
inline void SRT(T &A) {
  sort(A.begin(), A.end());
}
template <class T, class C>
inline void SRT(T &A, C B) {
  sort(A.begin(), A.end(), B);
}
const int MOD = 1000000007;
const int INF = 10009;
const double EPS = 1e-2;
const double OO = 1e15;
const double PI = 3.14159265358979323846264;
template <class T>
inline void checkMin(T &a, const T b) {
  if (b < a) a = b;
}
template <class T>
inline void checkMax(T &a, const T b) {
  if (b > a) a = b;
}
template <class T, class C>
inline void checkMin(T &a, const T b, C c) {
  if (c(b, a)) a = b;
}
template <class T, class C>
inline void checkMax(T &a, const T b, C c) {
  if (c(a, b)) a = b;
}
template <class T>
inline T min(T a, T b, T c) {
  return min(min(a, b), c);
}
template <class T>
inline T max(T a, T b, T c) {
  return max(max(a, b), c);
}
template <class T>
inline T min(T a, T b, T c, T d) {
  return min(min(a, b), min(c, d));
}
template <class T>
inline T max(T a, T b, T c, T d) {
  return max(min(a, b), max(c, d));
}
template <class T>
inline T sqr(T a) {
  return a * a;
}
template <class T>
inline T cub(T a) {
  return a * a * a;
}
int Ceil(int x, int y) { return (x - 1) / y + 1; }
inline bool _1(int x, int i) { return x & 1 << i; }
inline bool _1(long long x, int i) { return x & 1LL << i; }
inline long long _1(int i) { return 1LL << i; }
inline long long _U(int i) { return _1(i) - 1; };
inline int count_bits(int x) {
  x = (x & 0x55555555) + ((x & 0xaaaaaaaa) >> 1);
  x = (x & 0x33333333) + ((x & 0xcccccccc) >> 2);
  x = (x & 0x0f0f0f0f) + ((x & 0xf0f0f0f0) >> 4);
  x = (x & 0x00ff00ff) + ((x & 0xff00ff00) >> 8);
  x = (x & 0x0000ffff) + ((x & 0xffff0000) >> 16);
  return x;
}
template <class T>
inline T low_bit(T x) {
  return x & -x;
}
template <class T>
inline T high_bit(T x) {
  T p = low_bit(x);
  while (p != x) x -= p, p = low_bit(x);
  return p;
}
template <class T>
inline void RD(T &x) {
  scanf("%d", &x);
}
template <class T>
inline void OT(const T &p) {
  printf("%.0lfn", p);
}
const int N = 100009;
vector<int> adj[N];
int n, m, ans;
int main() {
  int a, b;
  RD(n, m);
  for (int i = 0; i < int(m); ++i) {
    RD(a, b), --a, --b, adj[a].push_back(b), adj[b].push_back(a);
  }
  for (int i = 0; i < int(n); ++i)
    if (int(adj[i].size()) & 1) ++ans;
  ans >>= 1;
  for (int i = 0; i < int(n); ++i)
    if (int(adj[i].size()) == 2) {
      a = adj[i][0], b = adj[i][1];
      if (a == b) {
        CLR(adj[a]);
        ++ans;
      } else {
        adj[a][adj[a][0] == i ? 0 : 1] = b;
        adj[b][adj[b][0] == i ? 0 : 1] = a;
      }
    }
  printf("%d %d\n", ans, m);
}