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

typedef long long ll;
const int N = 1e5 + 5;
vector<int> edge[N];
int col[N];
int a[3];
int n, m;
bool bp;

void DFS(int u) {
    for (int v: edge[u]) {
        if (col[v] != -1) {
            if ((col[v]^col[u]) != 1) bp = false;
        } else {
            col[v] = col[u]^1;
            DFS(v);
        }
    }
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i=1; i<=m; ++i) {
        int u, v;
        scanf("%d%d", &u, &v);
        edge[u].push_back(v);
        edge[v].push_back(u);
    }
    memset(col, -1, sizeof(col));
    for (int i=1; i<=n; ++i) {
        if (col[i] != -1) continue;
        if (edge[i].size() == 0) { col[i] = 0; a[0]++; }
        else {
            col[i] = 0;
            bp = true;
            DFS(i);
            if (bp) a[1]++;
            else a[2]++;
        }
    }
    ll ans = 0;
    for (int i=1; i<3; ++i) {
        for (int j=1; j<3; ++j) {
            if (i == 1 && j == 1) ans += (ll)a[i]*a[j]*2LL;
            else ans += (ll)a[i]*a[j];
        }
    }
    ans += (ll)a[0]*(n-a[0])*2LL+(ll)a[0]*a[0];
//    printf("%lld\n", ans);
    cout << ans << endl;
    return 0;
}