Problem: 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.

Examples

Input

4 3
1 3 2
2 4 3
1 4 6


Output

5


Input

4 2
1 2 1
3 4 2


Output

-1


Input

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


Output

28

Solution (in CPP):

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

#define dump(x) cout << (x) << '\n'
#define Int int64_t
#define fi first
#define se second

Int INF = 1e18;
//Int mod = 1e9+7;
Int mod = 998244353;

vector<pair<Int, Int>> G[100001];

int main() {
    Int n, m;
    cin >> n >> m;    
    for (Int i = 0; i < m; i++) {
        Int a, b, c;
        cin >> a >> b >> c;
        G[a].emplace_back(b, c);
    }
    for (Int i = 1; i < n; i++) {
        G[i + 1].emplace_back(i, 0);
    }
    vector<Int> d(n + 1, INF);
    d[1] = 0;
    priority_queue<pair<Int, Int>, vector<pair<Int,Int>>, greater<pair<Int, Int>>> q;
    q.push({0, 1});
    while (!q.empty()) {
        pair<Int, Int> p = q.top();
        q.pop();
        Int v = p.se;
        if (d[v] < p.fi) {
            continue;
        }
        for (Int i = 0; i < G[v].size(); i++) {
            pair<Int, Int> to = G[v][i];
            if (d[to.fi] > d[v] + to.se) {
                d[to.fi] = d[v] + to.se;
                q.push({d[to.fi], to.fi});
            }
        }
    }
    dump(d[n] == INF ? -1 : d[n]);
    return 0;
}