Problem:
Score : 800 points

Problem Statement

Takahashi has received an undirected graph with N vertices, numbered 1, 2, ..., N.
The edges in this graph are represented by (u_i, v_i).
There are no self-loops and multiple edges in this graph.

Based on this graph, Takahashi is now constructing a new graph with N^2 vertices, where each vertex is labeled with a pair of integers (a, b) (1 \leq a \leq N, 1 \leq b \leq N).
The edges in this new graph are generated by the following rule:

Span an edge between vertices (a, b) and (a', b') if and only if both of the following two edges exist in the original graph: an edge between vertices a and a', and an edge between vertices b and b'.

How many connected components are there in this new graph?

Constraints

2 \leq N \leq 100,000

0 \leq M \leq 200,000

1 \leq u_i < v_i \leq N

There exists no pair of distinct integers i and j such that u_i = u_j and v_i = v_j.

Input

The input is given from Standard Input in the following format:

N M
u_1 v_1
u_2 v_2
:
u_M v_M

Output

Print the number of the connected components in the graph constructed by Takahashi.

Sample Input 1

3 1
1 2

Sample Output 1

7

The graph constructed by Takahashi is as follows.

Sample Input 2

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

Sample Output 2

18

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

struct Edge {
  int t,next;
  Edge() {}
  Edge(int a,int b):t(a),next(b) {}
};

Edge e[400005];
int head[200005];

bool vis[100005],col[100005];
int cnt;
bool can;

void dfs(int x,int c) {
  vis[x]=1;cnt++;
  col[x]=c;
  for(int i=head[x];i;i=e[i].next)
    if (!vis[e[i].t]) dfs(e[i].t,c^1);
    else if (!(col[x]^col[e[i].t])) can=0;
}

int main() {
  int n,m;
  scanf("%d%d",&n,&m);
  for(int i=1;i<=m;i++) {
  	int x,y;
  	scanf("%d%d",&x,&y);
  	e[2*i-1]=Edge(y,head[x]);
  	head[x]=2*i-1;
  	e[2*i]=Edge(x,head[y]);
  	head[y]=2*i;
  }
  int s1=0,s2=0,s3=0;
  for(int i=1;i<=n;i++)
    if (!vis[i]) {
    	cnt=0;can=1;
    	dfs(i,0);
    	if (cnt==1) s3++;
    	else if (can) s1++;
    	else s2++;
	}
  ll ans=(ll)n*n-(ll)(n-s3)*(n-s3);
  ans+=(ll)(s1+s2)*(s1+s2)+(ll)s1*s1;
  printf("%lld\n",ans);
  return 0;
}