#include <fstream>
#include <vector>
using namespace std;
const int MAX_N = 200002;
const int MAX_M = 400002;
const int INF = (1 << 30);
typedef int Heap[MAX_N];
int N, M, ans;
int p[MAX_N], pos[MAX_N], val[MAX_N], sol[MAX_N][2];
vector < pair < int, int > > v[MAX_N];
bool m[MAX_N];
Heap H;
inline int left_son(int k) {
return 2*k;
}
inline int right_son(int k) {
return 2*k+1;
}
inline int father(int k) {
return k/2;
}
inline void sift(Heap H, int N, int K) {
int son;
do {
son = 0;
if(left_son(K) <= N) {
son = left_son(K);
if(right_son(K) <= N && val[H[right_son(K)]] < val[H[son]])
son = right_son(K);
}
if(val[H[son]] >= val[H[K]])
son = 0;
if(son) {
int temp = H[K];
H[K] = H[son], H[son] = temp;
pos[H[K]] = K, pos[H[son]] = son;
K = son;
}
} while(son);
}
inline void percolate(Heap H, int N, int K) {
int temp = H[K];
while(K > 1 && val[temp] < val[H[father(K)]]) {
H[K] = H[father(K)];
pos[H[K]] = K;
K = father(K);
}
H[K] = temp, pos[temp] = K;
}
inline void cut(Heap H, int &N, int K) {
H[K] = H[N], --N;
if(N+1 == K)
return;
pos[H[K]] = K;
if(K > 1 && val[H[K]] < val[H[father(K)]])
percolate(H, N, K);
else sift(H, N, K);
}
int main() {
ifstream f("apm.in");
ofstream g("apm.out");
f >> N >> M;
for(int i = 1, x, y, c; i <= M; ++i) {
f >> x >> y >> c;
v[x].push_back(make_pair(y, c)), v[y].push_back(make_pair(x, c));
}
int Nr = 0;
for(int i = 2; i <= N; ++i)
H[++Nr] = i, pos[i] = Nr, val[i] = INF;
m[1] = true;
for(size_t j = 0; j < v[1].size(); ++j) {
int y = v[1][j].first, c = v[1][j].second;
if(c < val[y]) {
val[y] = c, p[y] = 1;
cut(H, Nr, pos[y]);
H[++Nr] = y, pos[y] = Nr;
if(val[y] < val[H[father(Nr)]])
percolate(H, Nr, Nr);
}
}
for(int i = 1; i < N; ++i) {
sol[i][0] = H[1], sol[i][1] = p[H[1]];
ans += val[H[1]];
int x = H[1];
m[x] = true;
cut(H, Nr, 1);
for(size_t j = 0; j < v[x].size(); ++j) {
int y = v[x][j].first, c = v[x][j].second;
if(c < val[y] && m[y] == false) {
val[y] = c;
p[y] = x;
cut(H, Nr, pos[y]);
H[++Nr] = y, pos[y] = Nr;
if(val[y] < val[H[father(Nr)]])
percolate(H, Nr, Nr);
}
}
}
g << ans << '\n' << N-1 << '\n';
for(int i = 1; i < N; ++i)
g << sol[i][0] << ' ' << sol[i][1] << '\n';
f.close();
g.close();
return 0;
}