#include <bits/stdc++.h>
using namespace std;

vector<long long> dijkstra(vector<vector<pair<int,long long>>> &Graph,int start){
int N = Graph.size();
//O((V+E)logV) 一般最短経路魔法.
long long inf = 3e18;
    vector<bool> used(N);
    vector<long long> ret(N,inf);
    priority_queue<pair<long long,int>,vector<pair<long long,int>>,greater<pair<long long,int>>> Q;
    ret.at(start) = 0; Q.push({0,start});
    while(Q.size()){
        auto[nowd,pos] = Q.top(); Q.pop();
        if(used.at(pos)) continue;
        used.at(pos) = true;
        for(auto [to,w] : Graph.at(pos)){
            if(ret.at(to) > nowd+w){
                ret.at(to) = nowd+w;
                Q.push({ret.at(to),to});
            }
        }
    }
    return ret;
}
vector<long long> dijkstra2(vector<vector<pair<int,long long>>> &Graph,int start){
    int N = Graph.size();
    //O(V^2) 密グラフ専用.
    long long inf = 3e18;
    vector<bool> used(N);
    vector<long long> ret(N,inf);
    ret.at(start) = 0;
    while(true){
        long long nowd = inf,pos = -1;
        for(int i=0; i<N; i++){
            if(used.at(i)) continue;
            if(nowd > ret.at(i)) nowd = ret.at(i),pos = i;
        }
        if(pos == -1) break;
        used.at(pos) = true;
 
        for(auto [to,w] : Graph.at(pos)){if(ret.at(to) > nowd+w) ret.at(to) = nowd+w;}
    }
    return ret;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    long long N,M,C; cin >> N >> M >> C;
    vector<vector<pair<int,long long>>> Graph(N),Graph2(N*2);
    while(M--){
        int u,v,w; cin >> u >> v >> w,u--,v--;
        Graph.at(u).push_back({v,w+C});
        Graph.at(v).push_back({u,w+C});

        Graph2.at(u).push_back({v,w+C});
        Graph2.at(v).push_back({u,w+C});
        Graph2.at(u).push_back({v+N,C});
        Graph2.at(v).push_back({u+N,C});
        u += N,v += N;
        Graph2.at(u).push_back({v,w+C});
        Graph2.at(v).push_back({u,w+C});
    }

    auto dist1 = dijkstra(Graph,0);
    auto dist2 = dijkstra(Graph2,N-1);
    for(int i=1; i<N; i++){
        long long answer = dist1.at(N-1);
        answer = min(answer,dist1.at(i)+dist2.at(N+i));
        cout << answer << "\n";
    }
}