A rabbit Taro decided to hold a party and invite some friends as guests. He has n rabbit friends, and m pairs of rabbits are also friends with each other. Friendliness of each pair is expressed with a positive integer. If two rabbits are not friends, their friendliness is assumed to be 0.

When a rabbit is invited to the party, his satisfaction score is defined as the minimal friendliness with any other guests. The satisfaction of the party itself is defined as the sum of satisfaction score for all the guests.

To maximize satisfaction scores for the party, who should Taro invite? Write a program to calculate the maximal possible satisfaction score for the party.



Input

The first line of the input contains two integers, n and m (1 \leq n \leq 100, 0 \leq m \leq 100). The rabbits are numbered from 1 to n.

Each of the following m lines has three integers, u, v and f. u and v (1 \leq u, v \leq n, u \neq v, 1 \leq f \leq 1,000,000) stands for the rabbits' number, and f stands for their friendliness.

You may assume that the friendliness of a pair of rabbits will be given at most once.

Output

Output the maximal possible satisfaction score of the party in a line.

Examples

Input

3 3
1 2 3
2 3 1
3 1 2


Output

6


Input

2 1
1 2 5


Output

10


Input

1 0


Output

0


Input

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


Output

16
#define  _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES

#pragma comment (linker, "/STACK:526000000")

#include "bits/stdc++.h"

using namespace std;
typedef string::const_iterator State;
#define eps 1e-11L
#define MAX_MOD 1000000007LL
#define GYAKU 500000004LL

#define MOD 998244353LL
#define seg_size 262144
#define pb push_back
#define mp make_pair
typedef long long ll;
#define REP(a,b) for(long long (a) = 0;(a) < (b);++(a))
#define ALL(x) (x).begin(),(x).end()

void init() {
    iostream::sync_with_stdio(false);
    cout << fixed << setprecision(20);
}


unsigned long xor128() {
    static unsigned long x = 123456789, y = 362436069, z = 521288629, w = 88675123;
    unsigned long t = (x ^ (x << 11));
    x = y; y = z; z = w;
    return (w = (w ^ (w >> 19)) ^ (t ^ (t >> 8)));
}

//li chao tree by ei1333. It's public available.
// https://ei1333.github.io/luzhiled/snippets/structure/li-chao-tree.html
/*
template< typename T >
struct LiChaoTree {
    struct Line {
        T a, b;

        Line(T a, T b) : a(a), b(b) {}

        inline T get(T x) const { return a * x + b; }

        inline bool over(const Line& b, const T& x) const {
            return get(x) < b.get(x);
        }
    };

    vector< T > xs;
    vector< Line > seg;
    int sz;

    LiChaoTree(const vector< T >& x, T INF) : xs(x) {
        sz = 1;
        while (sz < xs.size()) sz <<= 1;
        while (xs.size() < sz) xs.push_back(xs.back() + 1);
        seg.assign(2 * sz - 1, Line(0, INF));
    }

    void update(Line& x, int k, int l, int r) {
        int mid = (l + r) >> 1;
        auto latte = x.over(seg[k], xs[l]), malta = x.over(seg[k], xs[mid]);
        if (malta) swap(seg[k], x);
        if (l + 1 >= r) return;
        else if (latte != malta) update(x, 2 * k + 1, l, mid);
        else update(x, 2 * k + 2, mid, r);
    }

    void update(T a, T b) { // ax+b
        Line l(a, b);
        update(l, 0, 0, sz);
    }

    T query(int k) { // xs[k]
        const T x = xs[k];
        k += sz - 1;
        T ret = seg[k].get(x);
        while (k > 0) {
            k = (k - 1) >> 1;
            ret = min(ret, seg[k].get(x));
        }
        return ret;
    }
};

#define int long long
void solve() {
    int n, m;
    cin >> n >> m;
    vector<vector<int>> landmarks;
    vector<vector<int>> costs;
    REP(i, n) {
        vector<int> tmp;
        REP(q, m) {
            int b;
            cin >> b;
            b = xor128() % 10000000;
            tmp.push_back(b);
        }
        landmarks.push_back(tmp);
    }
    REP(i, n) {

        vector<int> tmp;
        {
            REP(q, m) {
                int b;
                cin >> b;
                b = xor128() % 100000000;
                tmp.push_back(b);
            }
        }
        costs.push_back(tmp);
    }
    vector<LiChaoTree<int>> yoko; // size should be m
    REP(i, n) {
        vector<int> hoge(m);
        REP(q, m) {
            hoge[q] = q;
        }
        LiChaoTree<int> now(hoge, 1e18);//size should be n
        yoko.push_back(now);
    }
    vector<vector<int>> dp = landmarks;
    for (int q = 0; q < m; ++q) {
        vector<int> hoge(n);
        REP(i, n) {
            hoge[i] = i;
        }
        LiChaoTree<int> now(hoge, 1e18);//size should be n

        for (int i = 0; i < n; ++i) {
            int now_min = costs[0][0] * (i + q);
            if (q < 0) {
                int geko = yoko[i].query(q);
                now_min = min(now_min, geko);
            }
            if (i != 0) {
                int geko = now.query(i);
                now_min = min(now_min, geko);
            }
            dp[i][q] = now_min;
            yoko[i].update(costs[i][q], landmarks[i][q] + dp[i][q] - q * costs[i][q]);
            now.update(costs[i][q], landmarks[i][q] + dp[i][q] - i * costs[i][q]);
        }
    }

    cout << dp[n - 1][m - 1] << endl;
}
*/

#define int ll
int used[200];
vector<int> kouho;
set<pair<int,int>> already;
int dist[200][200];
int n, m;

int dfs(int now,pair<int,int> bit) {
    if (now < 50) {
        bit.first += (1LL << now);
    }
    else {
        bit.second += (1LL << (now - 50));
    }
    if (already.count(bit)) {
        return 0;
    }
    used[now] = 1;
    kouho.push_back(now);
    already.insert(bit);
    for (int i = 0; i < n; ++i) {
        if (used[i]) continue;
        int ok = 1;
        REP(q, kouho.size()) {
            if (dist[kouho[q]][i] == 0) {
                ok = 0;
                break;
            }
        }
        if (ok == 0) continue;
        dfs(i, bit);
    }
    kouho.pop_back();
    used[now] = 0;
    return 0;
}

void solve() {
    cin >> n >> m;
    REP(i, m) {
        int a, b, c;
        cin >> a >> b >> c;
        a--; b--;
        dist[a][b] = c;
        dist[b][a] = c;
    }
    int ans = 0;
    REP(i, n) {
        ans = max(ans, dfs(i, pair<int, int>{0, 0}));
   }
    for (auto x : already) {
        vector<int> huku;
        for (int q = 0; q < 51; ++q) {
            if ((1LL << q) & x.first) {
                huku.push_back(q);
            }
            if ((1LL << q) & x.second) {
                huku.push_back(q + 50);
            }
        }
        if (huku.size() == 1) continue;
        int tmp = 0;
        REP(q, huku.size()) {
            int hoge = 1e9;
            REP(j, huku.size()) {
                if (q == j) continue;
                hoge = min(hoge, dist[huku[q]][huku[j]]);
            }
            tmp += hoge;
        }
        ans = max(ans, tmp);
    }
    cout << ans << endl;
}
#undef int
int main() {
    init();
    solve();
}
