#include <bits/stdc++.h>
using namespace std;
const int N = 110;
#define INF 0x3f3f3f3f
/**
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
*/
int n, m;
int g[N][N];
int dis[N][N]; // dp结果数组
int path[N][N];
int ans[N];
int cnt;
int res = INF;
void floyd() {
    for (int k = 1; k <= n; k++) {
        // dp
        for (int i = 1; i < k; i++) {
            for (int j = i + 1; j < k; j++) {              // i,j,k序号由小到大
                if (res - dis[i][j] > g[i][k] + g[k][j]) { // 减法防溢出
                    res = dis[i][j] + g[i][k] + g[k][j];

                    int x = i, y = j;
                    cnt = 0; // 以前有过的路径也清空
                    while (x != y) {
                        ans[cnt++] = y;
                        y = path[i][y];
                    }
                    ans[cnt++] = x;
                    ans[cnt++] = k; // 序号最大的节点k
                }
            }
        }
        // floyd
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                if (dis[i][j] > dis[i][k] + dis[k][j]) {
                    dis[i][j] = dis[i][k] + dis[k][j];
                    path[i][j] = path[k][j]; // 这咋还和我理解的不一样呢？
                }
    }
}

int main() {
    while (cin >> n >> m) {
        // 邻接矩阵初始化
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++) {
                dis[i][j] = g[i][j] = INF;
                path[i][j] = i; // 这里也是不一样，需要思考与整理
            }

        // 读入边
        while (m--) {
            int a, b, c;
            cin >> a >> b >> c;
            g[a][b] = g[b][a] = min(g[a][b], c);
            dis[a][b] = dis[b][a] = g[a][b];
        }

        floyd();

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