Description



You are given undirected weighted graph. Find the length of the shortest cycle which starts from the vertex 1 and passes throught all the edges at least once. Graph may contain multiply edges between a pair of vertices and loops (edges from the vertex to itself).


Input



The first line of the input contains two integers n and m (1 ≤ n ≤ 15, 0 ≤ m ≤ 2000), n is the amount of vertices, and m is the amount of edges. Following m lines contain edges as a triples x, y, w (1 ≤ x, y ≤ n, 1 ≤ w ≤ 10000), x, y are edge endpoints, and wis the edge length.


Output



Output minimal cycle length or -1


Sample Input



Input 



3 3
1 2 1
2 3 1
3 1 1



Output



3



Input 


3 2
1 2 3
2 3 4



Output



14




要求是每条边至少走一遍的最小花费，先转变为一个欧拉回路问题，即在奇数度的节点上加边使其变为偶数，所有点都为偶数度即有欧拉回路，然后是加边操作，先用floyd预处理出最短路即可。需要注意判断是否可以走遍所有边。



#include<map>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<bitset>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int maxn = 20;
int T, n, m, ans, x, y, z;
int dis[maxn][maxn], cnt[maxn], fa[maxn];
int a[maxn], tot, vis[maxn];

int get(int x)
{
  return x == fa[x] ? x : fa[x] = get(fa[x]);
}

int dfs()
{
  for (int i = 0; i < tot; i++)
  {
    if (vis[i]) continue;
    int ans = INF;
    for (int j = i + 1; j < tot; j++)
    {
      if (vis[j]) continue;
      vis[i] = vis[j] = 1;
      ans = min(ans, dis[a[i]][a[j]] + dfs());
      vis[i] = vis[j] = 0;
    }
    return ans;
  }
  return 0;
}

int main()
{
  while (~scanf("%d%d", &n, &m))
  {
    ans = 0;
    for (int i = 1; i <= n; i++)
    {
      cnt[i] = 0; fa[i] = i;
      for (int j = 1; j <= n; j++) dis[i][j] = INF;
    }
    int flag1 = 0, flag2 = 0;
    for (int i = 1; i <= m; i++)
    {
      scanf("%d%d%d", &x, &y, &z);
      ans += z;
      if (++cnt[x] == 1) flag2++;
      if (++cnt[y] == 1) flag2++;
      int fx = get(x), fy = get(y);
      if (fx != fy) { fa[fx] = fy; flag1++; }
      dis[x][y] = min(dis[x][y], z);
      dis[y][x] = min(dis[y][x], z);
    }
    if (n == 1 || m == 0) { printf("%d\n", ans); continue; }
    if (flag1 + 1 != flag2 || cnt[1] == 0) { printf("-1\n"); continue; }
    for (int k = 1; k <= n; k++)
    {
      for (int i = 1; i <= n; i++)
      {
        if (k == i || dis[i][k] == INF) continue;
        for (int j = 1; j <= n; j++)
        {
          if (j == i || j == k) continue;
          if (dis[k][j] == INF) continue;
          if (dis[i][j] > dis[i][k] + dis[k][j])
          {
            dis[i][j] = dis[i][k] + dis[k][j];
          }
        }
      }
    }
    tot = 0;
    for (int i = 1; i <= n; i++)
    {
      if (cnt[i] & 1)
      {
        vis[tot] = 0;
        a[tot++] = i;
      }
    }
    printf("%d\n", ans + dfs());
  }
  return 0;
}