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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int N, M;
vector<int> adjlist[100005];
bool vis[100005], fail;
int c[100005];

void dfs(int x, int col) {
	if (vis[x]) {
		if (col != c[x]) fail = 1;
		return;
	} else vis[x] = 1;
	c[x] = col;
	for (int i = 0; i < adjlist[x].size(); i++) dfs(adjlist[x][i], !col);
		
}
long long l, r;

int main() {
	scanf("%d%d", &N, &M);
	for (int i = 0; i < M; i++) {
		int a, b;
		scanf("%d%d", &a, &b);
		a--;
		b--;
		adjlist[a].push_back(b);
		adjlist[b].push_back(a);
	}
	long long ans = 0;
	for (int i = 0; i < N; i++) {
		if (vis[i]) continue;
		if (adjlist[i].size() == 0) {
			ans += N * 2 - 1;
			continue;
		}
		fail = 0;
		dfs(i, 0);
		if (fail) l++;
		else r++;
	}
	//printf("%lld %lld %lld\n", ans, l, r);
	ans += l * l;
	ans += l * r * 2;
	ans += r * (r - 1) * 4;
	ans += r * 2;
	printf("%lld\n", ans);
}