#include <bits/stdc++.h>
#define ll long long
#define pll pair<ll, ll>
#define pii pair<int, int>
#define f first
#define se second
#define pb push_back
#define ld long double
 
using namespace std;
 
 
const int N = 2e5 + 123;
const int MAXN = 1e5 + 12;
const int inf = 1e9 + 7;
const ll mod = 1e9 + 7;
 

ll n, m, d[N];
bool vis[N], is[N];
vector <pll> g[N];
 
 
void djk() {
    priority_queue <pair<ll, ll> > q;
    q.push({0, 1});
    pair<ll, ll> tmp;
    ll u, w;
    while (!q.empty()) {
        tmp = q.top();
        q.pop();
        u = tmp.se;
        w = -tmp.f;
        if (vis[u])
            continue;
        vis[u] = 1;
        d[u] = w;
        if (u != 1 && !vis[u - 1])
            q.push({-w, u - 1});
        for (int i = 0; i < g[u].size(); i++)
            if (!vis[g[u][i].f])
                q.push({-w - g[u][i].se, g[u][i].f});
    }
} 
 
 
int main() {
    cin >> n >> m;
    ll x, y, z;
    memset(is, 1, sizeof(is));
    for (int i = 1; i <= m; i++) {
        cin >> x >> y >> z;
        g[x].pb({y, z});
    }
    memset(d, -1, sizeof(d));
    djk();
    cout << d[n];
    return 0;
}
### Prompt

In Cpp, your task is to solve the following 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

### Response

```cpp
#include <bits/stdc++.h>
#define ll long long
#define pll pair<ll, ll>
#define pii pair<int, int>
#define f first
#define se second
#define pb push_back
#define ld long double
 
using namespace std;
 
 
const int N = 2e5 + 123;
const int MAXN = 1e5 + 12;
const int inf = 1e9 + 7;
const ll mod = 1e9 + 7;
 

ll n, m, d[N];
bool vis[N], is[N];
vector <pll> g[N];
 
 
void djk() {
    priority_queue <pair<ll, ll> > q;
    q.push({0, 1});
    pair<ll, ll> tmp;
    ll u, w;
    while (!q.empty()) {
        tmp = q.top();
        q.pop();
        u = tmp.se;
        w = -tmp.f;
        if (vis[u])
            continue;
        vis[u] = 1;
        d[u] = w;
        if (u != 1 && !vis[u - 1])
            q.push({-w, u - 1});
        for (int i = 0; i < g[u].size(); i++)
            if (!vis[g[u][i].f])
                q.push({-w - g[u][i].se, g[u][i].f});
    }
} 
 
 
int main() {
    cin >> n >> m;
    ll x, y, z;
    memset(is, 1, sizeof(is));
    for (int i = 1; i <= m; i++) {
        cin >> x >> y >> z;
        g[x].pb({y, z});
    }
    memset(d, -1, sizeof(d));
    djk();
    cout << d[n];
    return 0;
}
```