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;
namespace fastIO {
bool IOerror = 0;
inline char nc() {
  static char buf[100005], *p1 = buf + 100005, *pend = buf + 100005;
  if (p1 == pend) {
    p1 = buf;
    pend = buf + fread(buf, 1, 100005, stdin);
    if (pend == p1) {
      IOerror = 1;
      return -1;
    }
  }
  return *p1++;
}
inline bool blank(char ch) {
  return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
}
inline void read(int &x) {
  bool sign = 0;
  char ch = nc();
  x = 0;
  for (; blank(ch); ch = nc())
    ;
  if (IOerror) return;
  if (ch == '-') sign = 1, ch = nc();
  for (; ch >= '0' && ch <= '9'; ch = nc()) x = x * 10 + ch - '0';
  if (sign) x = -x;
}
inline void read(long long &x) {
  bool sign = 0;
  char ch = nc();
  x = 0;
  for (; blank(ch); ch = nc())
    ;
  if (IOerror) return;
  if (ch == '-') sign = 1, ch = nc();
  for (; ch >= '0' && ch <= '9'; ch = nc()) x = x * 10 + ch - '0';
  if (sign) x = -x;
}
inline void read(double &x) {
  bool sign = 0;
  char ch = nc();
  x = 0;
  for (; blank(ch); ch = nc())
    ;
  if (IOerror) return;
  if (ch == '-') sign = 1, ch = nc();
  for (; ch >= '0' && ch <= '9'; ch = nc()) x = x * 10 + ch - '0';
  if (ch == '.') {
    double tmp = 1;
    ch = nc();
    for (; ch >= '0' && ch <= '9'; ch = nc())
      tmp /= 10.0, x += tmp * (ch - '0');
  }
  if (sign) x = -x;
}
inline void read(char *s) {
  char ch = nc();
  for (; blank(ch); ch = nc())
    ;
  if (IOerror) return;
  for (; !blank(ch) && !IOerror; ch = nc()) *s++ = ch;
  *s = 0;
}
inline void read(char &c) {
  for (c = nc(); blank(c); c = nc())
    ;
  if (IOerror) {
    c = -1;
    return;
  }
}
inline void read1(int &x) {
  char ch;
  int bo = 0;
  x = 0;
  for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') bo = 1;
  for (; ch >= '0' && ch <= '9'; x = x * 10 + ch - '0', ch = getchar())
    ;
  if (bo) x = -x;
}
inline void read1(long long &x) {
  char ch;
  int bo = 0;
  x = 0;
  for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') bo = 1;
  for (; ch >= '0' && ch <= '9'; x = x * 10 + ch - '0', ch = getchar())
    ;
  if (bo) x = -x;
}
inline void read1(double &x) {
  char ch;
  int bo = 0;
  x = 0;
  for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') bo = 1;
  for (; ch >= '0' && ch <= '9'; x = x * 10 + ch - '0', ch = getchar())
    ;
  if (ch == '.') {
    double tmp = 1;
    for (ch = getchar(); ch >= '0' && ch <= '9';
         tmp /= 10.0, x += tmp * (ch - '0'), ch = getchar())
      ;
  }
  if (bo) x = -x;
}
inline void read1(char *s) {
  char ch = getchar();
  for (; blank(ch); ch = getchar())
    ;
  for (; !blank(ch); ch = getchar()) *s++ = ch;
  *s = 0;
}
inline void read1(char &c) {
  for (c = getchar(); blank(c); c = getchar())
    ;
}
inline void read2(int &x) { scanf("%d", &x); }
inline void read2(long long &x) { scanf("%lld", &x); }
inline void read2(double &x) { scanf("%lf", &x); }
inline void read2(char *s) { scanf("%s", s); }
inline void read2(char &c) { scanf(" %c", &c); }
inline void readln2(char *s) { gets(s); }
struct Ostream_fwrite {
  char *buf, *p1, *pend;
  Ostream_fwrite() {
    buf = new char[100005];
    p1 = buf;
    pend = buf + 100005;
  }
  void out(char ch) {
    if (p1 == pend) {
      fwrite(buf, 1, 100005, stdout);
      p1 = buf;
    }
    *p1++ = ch;
  }
  void print(int x) {
    static char s[15], *s1;
    s1 = s;
    if (!x) *s1++ = '0';
    if (x < 0) out('-'), x = -x;
    while (x) *s1++ = x % 10 + '0', x /= 10;
    while (s1-- != s) out(*s1);
  }
  void println(int x) {
    static char s[15], *s1;
    s1 = s;
    if (!x) *s1++ = '0';
    if (x < 0) out('-'), x = -x;
    while (x) *s1++ = x % 10 + '0', x /= 10;
    while (s1-- != s) out(*s1);
    out('\n');
  }
  void print(long long x) {
    static char s[25], *s1;
    s1 = s;
    if (!x) *s1++ = '0';
    if (x < 0) out('-'), x = -x;
    while (x) *s1++ = x % 10 + '0', x /= 10;
    while (s1-- != s) out(*s1);
  }
  void println(long long x) {
    static char s[25], *s1;
    s1 = s;
    if (!x) *s1++ = '0';
    if (x < 0) out('-'), x = -x;
    while (x) *s1++ = x % 10 + '0', x /= 10;
    while (s1-- != s) out(*s1);
    out('\n');
  }
  void print(double x, int y) {
    static long long mul[] = {1,
                              10,
                              100,
                              1000,
                              10000,
                              100000,
                              1000000,
                              10000000,
                              100000000,
                              1000000000,
                              10000000000LL,
                              100000000000LL,
                              1000000000000LL,
                              10000000000000LL,
                              100000000000000LL,
                              1000000000000000LL,
                              10000000000000000LL,
                              100000000000000000LL};
    if (x < -1e-12) out('-'), x = -x;
    x *= mul[y];
    long long x1 = (long long)floor(x);
    if (x - floor(x) >= 0.5) ++x1;
    long long x2 = x1 / mul[y], x3 = x1 - x2 * mul[y];
    print(x2);
    if (y > 0) {
      out('.');
      for (size_t i = 1; i < y && x3 * mul[i] < mul[y]; out('0'), ++i)
        ;
      print(x3);
    }
  }
  void println(double x, int y) {
    print(x, y);
    out('\n');
  }
  void print(char *s) {
    while (*s) out(*s++);
  }
  void println(char *s) {
    while (*s) out(*s++);
    out('\n');
  }
  void flush() {
    if (p1 != buf) {
      fwrite(buf, 1, p1 - buf, stdout);
      p1 = buf;
    }
  }
  ~Ostream_fwrite() { flush(); }
} Ostream;
inline void print(int x) { Ostream.print(x); }
inline void println(int x) { Ostream.println(x); }
inline void print(char x) { Ostream.out(x); }
inline void println(char x) {
  Ostream.out(x);
  Ostream.out('\n');
}
inline void print(long long x) { Ostream.print(x); }
inline void println(long long x) { Ostream.println(x); }
inline void print(double x, int y) { Ostream.print(x, y); }
inline void println(double x, int y) { Ostream.println(x, y); }
inline void print(char *s) { Ostream.print(s); }
inline void println(char *s) { Ostream.println(s); }
inline void println() { Ostream.out('\n'); }
inline void flush() { Ostream.flush(); }
char Out[100005], *o = Out;
inline void print1(int x) {
  static char buf[15];
  char *p1 = buf;
  if (!x) *p1++ = '0';
  if (x < 0) *o++ = '-', x = -x;
  while (x) *p1++ = x % 10 + '0', x /= 10;
  while (p1-- != buf) *o++ = *p1;
}
inline void println1(int x) {
  print1(x);
  *o++ = '\n';
}
inline void print1(long long x) {
  static char buf[25];
  char *p1 = buf;
  if (!x) *p1++ = '0';
  if (x < 0) *o++ = '-', x = -x;
  while (x) *p1++ = x % 10 + '0', x /= 10;
  while (p1-- != buf) *o++ = *p1;
}
inline void println1(long long x) {
  print1(x);
  *o++ = '\n';
}
inline void print1(char c) { *o++ = c; }
inline void println1(char c) {
  *o++ = c;
  *o++ = '\n';
}
inline void print1(char *s) {
  while (*s) *o++ = *s++;
}
inline void println1(char *s) {
  print1(s);
  *o++ = '\n';
}
inline void println1() { *o++ = '\n'; }
inline void flush1() {
  if (o != Out) {
    if (*(o - 1) == '\n') *--o = 0;
    puts(Out);
  }
}
struct puts_write {
  ~puts_write() { flush1(); }
} _puts;
inline void print2(int x) { printf("%d", x); }
inline void println2(int x) { printf("%d\n", x); }
inline void print2(char x) { printf("%c", x); }
inline void println2(char x) { printf("%c\n", x); }
inline void print2(long long x) { printf("%lld", x); }
inline void println2(long long x) {
  print2(x);
  printf("\n");
}
inline void println2() { printf("\n"); }
};  // namespace fastIO
using namespace fastIO;
vector<int> r[101000];
int n, m;
vector<int> c[101000];
int fa[101000], fl[101000], ans = 1, pre[101000], used[101000];
int GF(int x) { return x == fa[x] ? x : fa[x] = GF(fa[x]); }
void dfs(int x, int fff) {
  int y;
  for (int i = 0; i < (int)r[x].size(); i++) {
    y = r[x][i];
    if (y == fff) continue;
    if (!used[y]) {
      pre[y] = x;
      used[y] = used[x] + 1;
      dfs(y, x);
    } else if (used[y] < used[x]) {
      for (int t = x; t != y; t = pre[t]) {
        fl[GF(y)] = 1;
        fa[GF(t)] = GF(y);
      }
    }
  }
}
void doit(vector<int> &the) {
  int tot = 0, d = 0, t = 0, els = 0, x, y;
  for (int nnd = 0; nnd < (int)the.size(); nnd++) {
    x = the[nnd];
    t = 0;
    for (int i = 0; i < (int)r[x].size(); i++) {
      y = r[x][i];
      if (fa[x] != fa[y]) {
        t++;
        break;
      }
    }
    d += !!t;
  }
  if (d < 2) ans++;
}
int main() {
  read(n);
  read(m);
  int x, y;
  for (int i = 1; i <= m; i++) {
    read(x);
    read(y);
    r[x].push_back(y);
    r[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) fa[i] = i, ans += r[i].size() % 2;
  used[1] = 1;
  dfs(1, 0);
  for (int i = 1; i <= n; i++) c[GF(i)].push_back(i);
  ans /= 2;
  for (int i = 1; i <= n; i++)
    if (fa[i] == i)
      if (fl[i]) doit(c[i]);
  print(ans);
  print(' ');
  print(m);
}