Problem: You are given an undirected graph G. G has N vertices and M edges. The vertices are numbered from 1 through N, and the i-th edge (1 ≤ i ≤ M) connects Vertex a_i and b_i. G does not have self-loops and multiple edges.

You can repeatedly perform the operation of adding an edge between two vertices. However, G must not have self-loops or multiple edges as the result. Also, if Vertex 1 and 2 are connected directly or indirectly by edges, your body will be exposed to a voltage of 1000000007 volts. This must also be avoided.

Under these conditions, at most how many edges can you add? Note that Vertex 1 and 2 are never connected directly or indirectly in the beginning.

Constraints

* 2 ≤ N ≤ 10^5
* 0 ≤ M ≤ 10^5
* 1 ≤ a_i < b_i ≤ N
* All pairs (a_i, b_i) are distinct.
* Vertex 1 and 2 in G are not connected directly or indirectly by edges.

Input

Input is given from Standard Input in the following format:


N M
a_1 b_1
:
a_M b_M


Output

Print the maximum number of edges that can be added.

Examples

Input

4 1
1 3


Output

2


Input

2 0


Output

0


Input

9 6
1 4
1 8
2 5
3 6
4 8
5 7


Output

12

Solution (in CPP):

Solution:
#include "iostream"
#include "climits"
#include "list"
#include "queue"
#include "stack"
#include "set"
#include "functional"
#include "algorithm"
#include "string"
#include "map"
#include "unordered_map"
#include "unordered_set"
#include "iomanip"
#include "cmath"
#include "random"
#include "bitset"
#include "cstdio"
#include "numeric"
#include "cassert"
#include "ctime"

using namespace std;

constexpr long long int MOD = 1000000007;
//constexpr int MOD = 1000000007;
//constexpr int MOD = 998244353;
//constexpr long long int MOD = 998244353;
constexpr double EPS = 1e-8;

//int N, M, K, H, W, L, R;
long long int N, M, K, H, W, L, R;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	cin >> N >> M;
	vector<vector<int>>edge(N);
	for (int i = 0; i < M; i++) {
		cin >> L >> R;
		L--, R--;
		edge[L].push_back(R);
		edge[R].push_back(L);
	}
	long long int a = 1, b = 1;
	vector<int>used(N);
	for (int i = 0; i < 2; i++) {
		used[i] = 1;
		queue<int>Q;
		Q.push(i);
		while (!Q.empty()) {
			int cn = Q.front();
			Q.pop();
			for (auto j : edge[cn]) {
				if (used[j])continue;
				used[j] = 1;
				if (i == 0)a++;
				else b++;
				Q.push(j);
			}
		}
	}
	if (a < b)swap(a, b);
	a = N - b;
	cout << a * (a - 1) / 2 + b * (b - 1) / 2 - M << endl;
}