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 MAXN 100005
#define INF 1000000000000000000LL
#define MOD 998244353
#define F first
#define S second
#define exit dskosad
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
int n,m,l[MAXN],r[MAXN],c[MAXN];
vector<P> enter[MAXN];
vector<ll> exit[MAXN];
multiset<ll> ms;
ll dp[MAXN];
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d%d",&l[i],&r[i],&c[i]);
        enter[l[i]].push_back(P(r[i],c[i]));
    }
    dp[1]=0;
    for(int i=1;i<=n;i++)
    {
        if(i!=1)
        {
            if(ms.size()) dp[i]=*ms.begin();
            else dp[i]=INF;
        }
        for(auto p:enter[i])
        {
            ll res=dp[i]+p.S;
            ms.insert(res);
            exit[p.F].push_back(res);
        }
        for(auto x:exit[i]) ms.erase(ms.find(x));
    }
    if(dp[n]>=INF) puts("-1"); else printf("%lld\n",dp[n]);
    return 0;
}