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>
#define ll long long
#define ff first
#define ss second
#define mp make_pair
#define pb push_back
#define M 998244353
using namespace std;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll n,m,i,j,c;
    cin>>n>>m;
    pair <pair<ll,ll> ,ll> q[m];
    for(i=0;i<m;i++)
        cin>>q[i].ff.ff>>q[i].ff.ss>>q[i].ss;
    sort(q,q+m);
    
    multiset <pair<ll,ll>> ms;
    multiset <ll> cst;
    j = 0;
    cst.insert(0);
    ms.insert(mp(1,0));
    for(i=1;i<=n;i++)
    {
        if(cst.size() == 0)
        {
            cout<<-1;
            return 0;
        }
        if(i == n)
        {
            cout<<*cst.begin();
            return 0;
        } 
        while(j<m and q[j].ff.ff == i)
        {
            c = *cst.begin();
            cst.insert(c+q[j].ss);
            ms.insert(mp(q[j].ff.ss, c+q[j].ss));
            j++;
        }
        while(ms.size() and ms.begin()->ff == i)
        {
            cst.erase(cst.find(ms.begin()->ss));
            ms.erase(ms.begin());
        }

    }
    return 0;
}