Problem:
Score : 600 points

Problem Statement

We have N points numbered 1 to N arranged in a line in this order.

Takahashi decides to make an undirected graph, using these points as the vertices.
In the beginning, the graph has no edge. Takahashi will do M operations to add edges in this graph.
The i-th operation is as follows:

The operation uses integers L_i and R_i between 1 and N (inclusive), and a positive integer C_i. For every pair of integers (s, t) such that L_i \leq s < t \leq R_i, add an edge of length C_i between Vertex s and Vertex t.

The integers L_1, ..., L_M, R_1, ..., R_M, C_1, ..., C_M are all given as input.

Takahashi wants to solve the shortest path problem in the final graph obtained. Find the length of the shortest path from Vertex 1 to Vertex N in the final graph.

Constraints

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq L_i < R_i \leq N

1 \leq C_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N M
L_1 R_1 C_1
:
L_M R_M C_M

Output

Print the length of the shortest path from Vertex 1 to Vertex N in the final graph.
If there is no shortest path, print -1 instead.

Sample Input 1

4 3
1 3 2
2 4 3
1 4 6

Sample Output 1

5

We have an edge of length 2 between Vertex 1 and Vertex 2, and an edge of length 3 between Vertex 2 and Vertex 4, so there is a path of length 5 between Vertex 1 and Vertex 4.

Sample Input 2

4 2
1 2 1
3 4 2

Sample Output 2

-1

Sample Input 3

10 7
1 5 18
3 4 8
1 3 5
4 7 10
5 9 8
6 10 5
8 10 3

Sample Output 3

28

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using lint = long long int;
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
constexpr int MOD = 1000000007;
int main()
{
    int N, M;
    cin >> N >> M;
    vector<tuple<int, int, lint>> pair(M);
    REP(i, M) {
        int L, R;
        lint C;
        cin >> L >> R;
        cin >> C;
        pair[i] = make_tuple(L-1, R-1, C);
    }
    sort(pair.begin(), pair.end(),
       [](auto const &a, auto const &b) { return get<0>(a) < get<0>(b); });
    vector<lint> dp(N, 1e15);
    priority_queue<lint, vector<lint>, greater<lint> > pq;
    dp[0] = 0;
    int i = 0;
    REP(k, N) {
        int L = get<0>(pair[i]);
        if(L > k) {
            k++;
            if(pq.empty()) {
                cout << -1 << endl;
                return 0;
            }
            lint tmp = pq.top();
            if(dp[k] > tmp) {
                lint min = 1e15;
                int idx = -1;
                FOR(j, k+1, N) {
                    if(dp[j] < min) {
                        min = dp[j];
                        idx = j;
                    }
                    
                } 
                if(idx == -1) {
                    cout << -1 << endl;
                    return 0;
                }
                FOR(j, k, idx) {
                    dp[j] = min;
                }
                pq.pop();
            } else if(dp[k] == tmp) {
                pq.pop();
            }

            k--;
            continue;
        }
        //cout << dp[L]+get<2>(pair[i]) << " " << dp[get<1>(pair[i])] << endl;
        if(dp[L]+get<2>(pair[i]) < dp[get<1>(pair[i])]) {
            dp[get<1>(pair[i])] = dp[L]+get<2>(pair[i]);
            pq.push(dp[L]+get<2>(pair[i]));
        }
        k--;
        i++;
        if(i == M) break;
    }
    if(dp[N-1] == 1e15) {
        cout << -1 << endl;
    } else {
        cout << dp[N-1] << endl;
    }
}