Problem: You are given an undirected graph with n vertices and m edges. You have to write a number on each vertex of this graph, each number should be either 0 or 1. After that, you write a number on each edge equal to the sum of numbers on vertices incident to that edge.

You have to choose the numbers you will write on the vertices so that there is at least one edge with 0 written on it, at least one edge with 1 and at least one edge with 2. How many ways are there to do it? Two ways to choose numbers are different if there exists at least one vertex which has different numbers written on it in these two ways.

Input

The first line contains two integers n and m (1 ≤ n ≤ 40, 0 ≤ m ≤ (n(n - 1))/(2)) — the number of vertices and the number of edges, respectively.

Then m lines follow, each line contains two numbers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) — the endpoints of the i-th edge. It is guaranteed that each pair of vertices is connected by at most one edge.

Output

Print one integer — the number of ways to write numbers on all vertices so that there exists at least one edge with 0 written on it, at least one edge with 1 and at least one edge with 2.

Examples

Input


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


Output


20


Input


4 4
1 2
2 3
3 4
4 1


Output


4


Input


35 29
1 2
2 3
3 1
4 1
4 2
3 4
7 8
8 9
9 10
10 7
11 12
12 13
13 14
14 15
15 16
16 17
17 18
18 19
19 20
20 21
21 22
22 23
23 24
24 25
25 26
26 27
27 28
28 29
29 30


Output


34201047040

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline bool upmin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <typename T>
inline bool upmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
const long double eps = 1e-9;
const long double pi = acos(-1);
const int oo = 1 << 30;
const long long loo = 1ll << 62;
const int mods = 1e9 + 7;
const int MAXN = 100005;
const int INF = 0x3f3f3f3f;
namespace FastIO {
constexpr int SIZE = (1 << 21) + 1;
int num = 0, f;
char ibuf[SIZE], obuf[SIZE], que[65], *iS, *iT, *oS = obuf,
                                                *oT = obuf + SIZE - 1, c;
inline void flush() {
  fwrite(obuf, 1, oS - obuf, stdout);
  oS = obuf;
}
inline void putc(char c) {
  *oS++ = c;
  if (oS == oT) flush();
}
inline void getc(char &c) {
  for (c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                       (iS == iT ? EOF : *iS++))
                     : *iS++);
       c == ' ' || c == '\n';
       c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                       (iS == iT ? EOF : *iS++))
                     : *iS++))
    ;
}
inline void reads(char *st) {
  int n;
  getc(st[n = 1]);
  for (c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                       (iS == iT ? EOF : *iS++))
                     : *iS++);
       c != ' ' && c != '\n' && c != EOF;
       c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                       (iS == iT ? EOF : *iS++))
                     : *iS++))
    st[++n] = c;
}
template <class I>
inline void read(I &x) {
  for (f = 1, c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                              (iS == iT ? EOF : *iS++))
                            : *iS++);
       c < '0' || c > '9';
       c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                       (iS == iT ? EOF : *iS++))
                     : *iS++))
    if (c == '-') f = -1;
  for (x = 0; c >= '0' && c <= '9';
       c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                       (iS == iT ? EOF : *iS++))
                     : *iS++))
    x = (x << 3) + (x << 1) + (c & 15);
  x *= f;
}
template <class I>
inline void print(I x) {
  if (x < 0) putc('-'), x = -x;
  if (!x) putc('0');
  while (x) que[++num] = x % 10 + 48, x /= 10;
  while (num) putc(que[num--]);
}
struct Flusher_ {
  ~Flusher_() { flush(); }
} io_Flusher_;
}  // namespace FastIO
using FastIO ::print;
using FastIO ::putc;
using FastIO ::read;
vector<int> e[50];
int col[50], vis[50], f[50], cnt[50], n, m;
int find(int x) { return f[x] == x ? f[x] : f[x] = find(f[x]); }
long long quick_pow(long long x, long long y) {
  long long ret = 1;
  for (; y; y >>= 1) {
    if (y & 1) ret = ret * x;
    x = x * x;
  }
  return ret;
}
int dfs(int x, int c) {
  int flag = 1;
  col[x] = c, vis[x] = 1;
  for (auto v : e[x]) {
    if (!vis[v])
      flag &= dfs(v, c ^ 1);
    else if (col[x] == col[v])
      flag = 0;
  }
  return flag;
}
int lk[50], LK[50], g[1 << 20];
long long solve() {
  int nl = n >> 1, nr = n - nl;
  for (int i = 0; i < nl; ++i) lk[i] = LK[i] = 0;
  for (int i = 0; i < nl; ++i)
    for (auto v : e[i + 1])
      if (v <= nl) lk[i] |= 1 << (v - 1);
  for (int i = 0; i < 1 << nl; ++i) g[i] = 1;
  for (int i = 0; i < 1 << nl; ++i)
    for (int j = 0; j < nl; ++j)
      if (((i >> j) & 1) && (lk[j] & i)) {
        g[i] = 0;
        break;
      }
  for (int i = 0; i < nl; ++i)
    for (int j = (1 << nl) - 1; j >= 0; --j)
      if (!((j >> i) & 1)) g[j | (1 << i)] += g[j];
  for (int i = 0; i < nr; ++i) lk[i] = 0;
  for (int i = 0; i < nr; ++i) {
    for (auto v : e[i + nl + 1])
      if (v > nl)
        lk[i] |= 1 << (v - nl - 1);
      else
        LK[i] |= 1 << (v - 1);
  }
  long long ans = 0;
  for (int i = 0; i < 1 << nr; ++i) {
    int flag = 1, t = 0;
    for (int j = 0; j < nr; ++j)
      if (((i >> j) & 1)) {
        if (lk[j] & i) {
          flag = 0;
          break;
        }
        t |= LK[j];
      }
    if (!flag) continue;
    ans += g[((1 << nl) - 1) ^ t];
  }
  return ans;
}
signed main() {
  read(n), read(m);
  if (n == 1) {
    print(0);
    return 0;
  }
  for (int i = 1; i <= n; ++i) f[i] = i, cnt[i] = 1;
  for (int i = 1, u, v; i <= m; ++i) {
    read(u), read(v), e[u].push_back(v), e[v].push_back(u);
    u = find(u), v = find(v);
    if (u == v) continue;
    f[u] = v, cnt[v] += cnt[u];
  }
  long long ans = quick_pow(2, n);
  ans -= solve() * 2;
  int num = 0;
  for (int i = 1; i <= n; ++i)
    if (f[i] == i) ++num;
  ans -= quick_pow(2, num);
  int Num = 0;
  for (int i = 1; i <= n; ++i)
    if (f[i] == i && cnt[i] == 1) ++Num;
  ans += quick_pow(2, Num + 1);
  int flag = 1;
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) flag &= dfs(i, 0);
  if (flag) ans += quick_pow(2, num);
  if (!m) ans -= quick_pow(2, n);
  print(ans), putc('\n');
  return 0;
}