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: Accepted)

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

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<pll, ll> tpl;

const int MAXN = 1e5 + 10;
const ll inf = 1e18;

int N, M;
ll L[MAXN], R[MAXN], C[MAXN];
ll D[MAXN];
ll F[MAXN];

ll SEG[MAXN * 4], LAZY[MAXN * 4];

void f(int here, int l, int r){
    SEG[here] = min(SEG[here], LAZY[here]);
    if(l != r){
        LAZY[here * 2] = min(LAZY[here * 2], LAZY[here]);
        LAZY[here * 2 + 1] = min(LAZY[here * 2 + 1], LAZY[here]);
    }
    LAZY[here] = inf;
}

void init(int here, int l, int r){
    SEG[here] = inf, LAZY[here] = inf;
    if(l == r) return;
    int mid = (l + r) / 2;
    init(here * 2, l, mid); init(here * 2 + 1, mid + 1, r);
}

void update(int here, int l, int r, int x, int y, ll val){
    f(here, l, r);
    if(y < l || r < x) return ;
    if(x <= l && r <= y) {
        LAZY[here] = val;
        f(here, l, r);
        return;
    }

    int mid = (l + r) / 2;
    update(here * 2, l, mid, x, y, val);    update(here * 2 + 1, mid + 1, r, x, y, val);
    SEG[here] = min(SEG[here * 2], SEG[here * 2 + 1]);
}

ll query(int here, int l, int r, int x, int y){
    f(here, l, r);
    if(y < l || r < x) return inf;
    if(x <= l && r <= y) return SEG[here];
    int mid = (l + r) / 2;
    ll res = query(here * 2, l, mid, x, y);
    res = min(res, query(here * 2 + 1, mid + 1, r, x, y));
    return res;
}

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

    cin >> N >> M;
    for(int i = 1 ; i <= M ; i++) cin >> L[i] >> R[i] >> C[i];

    vector<tpl> V;
    for(int i = 1 ; i <= M ; i++){
        V.push_back(tpl(pll(L[i], R[i]), C[i]));
    }   

    sort(V.begin(), V.end());

    init(1, 1, N);
    update(1, 1, N, 1, 1, 0);
    for(auto val : V){
        int l = val.first.first, r = val.first.second;
        ll v = val.second;
        ll b = query(1, 1, N, l, l);
        update(1, 1, N, l, r, b + v);
    }

    ll ans = query(1, 1, N, N, N);
    if(ans == inf) ans = -1;
    cout << ans;
    return 0;
}