#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
const int N=1e5+5;
int n,m;
typedef long long ll;
typedef std::pair<ll,int> P;
std::priority_queue<P,std::vector<P>,std::greater<P> > q;
std::vector<P> g[N];
ll dp[N];
const ll INF=1e18;
int main() {
    scanf("%d%d",&n,&m);
    for(int l,r,c;m--;) {
        scanf("%d%d%d",&l,&r,&c);
        g[l].push_back(P(c,r));
    }
    dp[1]=0;
    for(P c:g[1]) q.push(c);
    for(int i=2;i<=n;i++) {
        while(!q.empty()&&q.top().second<i) q.pop();
        if(q.empty()) {
            dp[i]=INF;
        }
        else {
            dp[i]=q.top().first;
            for(P c:g[i]) q.push(P(dp[i]+c.first,c.second));
        }
    }
    printf("%lld",dp[n]<INF?dp[n]:-1LL);
    return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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

### Response

```cpp
#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
const int N=1e5+5;
int n,m;
typedef long long ll;
typedef std::pair<ll,int> P;
std::priority_queue<P,std::vector<P>,std::greater<P> > q;
std::vector<P> g[N];
ll dp[N];
const ll INF=1e18;
int main() {
    scanf("%d%d",&n,&m);
    for(int l,r,c;m--;) {
        scanf("%d%d%d",&l,&r,&c);
        g[l].push_back(P(c,r));
    }
    dp[1]=0;
    for(P c:g[1]) q.push(c);
    for(int i=2;i<=n;i++) {
        while(!q.empty()&&q.top().second<i) q.pop();
        if(q.empty()) {
            dp[i]=INF;
        }
        else {
            dp[i]=q.top().first;
            for(P c:g[i]) q.push(P(dp[i]+c.first,c.second));
        }
    }
    printf("%lld",dp[n]<INF?dp[n]:-1LL);
    return 0;
}

```