#include <bits/stdc++.h>
using namespace std;
const int N = 110;
#define INF 0x7ffffff
/**
分析:模板题，理解floyd 的在 i , j  路径中没有包含k（因为此时k未用来更新），即可写出最小环
在INF这里wa了两发，习惯值 1e9 不是最大值（查找时判断一下可以避免，但。。。懒了）
*/
int n, m;
int mp[N][N];
int g[N][N];
int path[N][N];
int ans[N];
int cnt;
int mm;
void floyd() {
    mm = INF;
    for (int k = 1; k <= n; k++) {
        for (int i = 1; i < k; i++) {
            for (int j = i + 1; j < k; j++) {
                int x = g[i][j] + mp[k][i] + mp[k][j];
                if (x < mm) {
                    mm = x;
                    int tmp = j;
                    cnt = 0;

                    while (tmp != i) {
                        ans[cnt++] = tmp;
                        tmp = path[i][tmp];
                    }
                    ans[cnt++] = i;
                    ans[cnt++] = k;
                }
            }
        }

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                if (g[i][j] > g[i][k] + g[k][j]) {
                    g[i][j] = g[i][k] + g[k][j];
                    path[i][j] = path[k][j];
                }
            }
        }
    }
}

int main() {
    while (cin >> n >> m) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                g[i][j] = mp[i][j] = INF;
                path[i][j] = i;
            }
        }

        while (m--) {
            int a, b, c;
            cin >> a >> b >> c;
            g[a][b] = g[b][a] = min(g[a][b], c); // 防重边
            mp[a][b] = mp[b][a] = g[a][b];
        }

        floyd();

        if (mm == INF) {
            puts("No solution.");
            continue;
        }
        for (int i = 0; i < cnt; ++i) printf("%d%s", ans[i], (i == cnt - 1) ? "\n" : " ");
    }
    return 0;
}