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:
/*
    Mbak Rose Dong
*/
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
int l[100100], r[100100];
ll c[100100];
ll dp[100100];
bool pend[100100];
vector<int> in[100100], out[100100];
int main()
{
    int n,i,j,k,x,y,ans,m,q,curr,temp;
    scanf("%d %d",&n,&m);
    for(i=1;i<=n;i++) dp[i] = 1e16;
    for(i=1;i<=m;i++)
    {
        scanf("%d %d %lld",&l[i], &r[i], &c[i]);
        in[l[i]].push_back(i);
        out[r[i]].push_back(i);
    }
    set<pair<ll,int> > s;
    for(i=1;i<=n;i++)
    {
        while(!s.empty() && pend[(*s.begin()).second] == 1) s.erase(s.begin());
        if(!s.empty())
        {
            curr = (*s.begin()).second;
            ll cost = (*s.begin()).first;
            dp[i] = min(dp[i], cost);
        }
        for(auto x : in[i])
        {
            ll aa = 0;
            if(l[x] != 1) aa = dp[l[x]];
            s.insert({c[x] + aa, x});
        }
        for(auto x : out[i])
        {
            pend[x] = 1;
        }
    }
    if(dp[n] == 1e16) dp[n] = -1;
    printf("%lld\n", dp[n]);
    return 0;
}