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;
#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=n-1;i>=0;i--)
#define perl(i,r,l) for(ll i=r-1;i>=l;i--)
#define fi first
#define se second
#define pb push_back
#define ins insert
#define pqueue(x) priority_queue<x,vector<x>,greater<x>>
#define all(x) (x).begin(),(x).end()
#define CST(x) cout<<fixed<<setprecision(x)
#define vtpl(x,y,z) vector<tuple<x,y,z>>
#define rev(x) reverse(x);
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
using pl=pair<ll,ll>;
using vpl=vector<pl>;
using vvpl=vector<vpl>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9+10;
const ll INF=4e18;
const ll dy[8]={1,0,-1,0,1,1,-1,-1};
const ll dx[8]={0,-1,0,1,1,-1,1,-1};
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
template <class T>
class SegTree {
    int n;// 葉の数
    vector<T> data;// データを格納するvector
    T def; // 初期値かつ単位元
    function<T(T, T)> operation; // 区間クエリで使う処理
    function<T(T, T)> update;// 点更新で使う処理

    // 区間[a,b)の総和。ノードk=[l,r)に着目している。
    T _query(int a, int b, int k, int l, int r) {
        if (r <= a || b <= l) return def; // 交差しない
        if (a <= l && r <= b)
            return data[k]; // a,l,r,bの順で完全に含まれる
        else {
            T c1 = _query(a, b, 2 * k + 1, l, (l + r) / 2); // 左の子
            T c2 = _query(a, b, 2 * k + 2, (l + r) / 2, r); // 右の子
            return operation(c1, c2);
        }
    }

    public:
    // _n:必要サイズ, _def:初期値かつ単位元, _operation:クエリ関数,
    // _update:更新関数
    SegTree(size_t _n, T _def, function<T(T, T)> _operation,
        function<T(T, T)> _update)
        : def(_def), operation(_operation), update(_update) {
        n = 1;
        while (n < _n) {
            n *= 2;
        }
        data = vector<T>(2 * n - 1, def);
    }

    // 場所i(0-indexed)の値をxで更新
    void change(int i, T x) {
        i += n - 1;
        data[i] = update(data[i], x);
        while (i > 0) {
            i = (i - 1) / 2;
            data[i] = operation(data[i * 2 + 1], data[i * 2 + 2]);
        }
    }

    // [a, b)の区間クエリを実行
    T query(int a, int b) {
        return _query(a, b, 0, 0, n);
    }

    // 添字でアクセス
    T operator[](int i) {
        return data[i + n - 1];
    }
};
int main(){
    ll n,m;cin >> n >>m;
    vector<pair<pl,ll>> e(m);
    rep(i,m){
        cin >> e[i].fi.fi >> e[i].fi.se >> e[i].se;
        e[i].fi.fi--;e[i].fi.se--;
    }
    sort(all(e));
    SegTree<ll> st(n,inf,[](ll a,ll b){return min(a,b);},[](ll a,ll b){return min(a,b);});
    st.change(0,0);
    rep(i,m){
        ll cost=st.query(e[i].fi.fi,e[i].fi.se+1);
        st.change(e[i].fi.se,cost+e[i].se);
        /*rep(i,n)cout << st[i] <<" ";
        cout << endl;*/
    }
    //rep(i,n)cout << st[i] <<endl;
    if(st[n-1]==inf)cout << -1 <<endl;
    else cout << st[n-1] <<endl;
}