#include <bits/stdc++.h>

using namespace std;
#define int long long
#define x first
#define y second

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;

mt19937 rand1(time(0));
int randlr(int l, int r) {
    return rand1() % (r - l + 1) + l;
}

vector<pii> g[100009];
int dist[100009];

const int inf = 1e17;

signed main() {
    ios :: sync_with_stdio(0);
    cin.tie(0);

    int n, m;
    cin >> n >> m;

    while (m--) {
        int l, r, c;
        cin >> l >> r >> c;

        g[l - 1].push_back({r - 1, c});
    }

    for (int i = 1; i < n; i ++) {
        g[i].push_back({i - 1, 0});
        dist[i] = inf;
    }

    priority_queue<pii> q;
    q.push({0, 0});
    while (q.size()) {
        auto vp = q.top();
        q.pop();

        if (-vp.x != dist[vp.y]) continue;
        int v = vp.y;

        for (auto u : g[v]) {
            if (dist[u.x] > dist[v] + u.y) {
                dist[u.x] = dist[v] + u.y;
                q.push({-dist[u.x], u.x});
            }
        }
    }

    if (dist[n - 1] == inf) dist[n - 1] = -1;

    cout << dist[n - 1];

    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>

using namespace std;
#define int long long
#define x first
#define y second

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;

mt19937 rand1(time(0));
int randlr(int l, int r) {
    return rand1() % (r - l + 1) + l;
}

vector<pii> g[100009];
int dist[100009];

const int inf = 1e17;

signed main() {
    ios :: sync_with_stdio(0);
    cin.tie(0);

    int n, m;
    cin >> n >> m;

    while (m--) {
        int l, r, c;
        cin >> l >> r >> c;

        g[l - 1].push_back({r - 1, c});
    }

    for (int i = 1; i < n; i ++) {
        g[i].push_back({i - 1, 0});
        dist[i] = inf;
    }

    priority_queue<pii> q;
    q.push({0, 0});
    while (q.size()) {
        auto vp = q.top();
        q.pop();

        if (-vp.x != dist[vp.y]) continue;
        int v = vp.y;

        for (auto u : g[v]) {
            if (dist[u.x] > dist[v] + u.y) {
                dist[u.x] = dist[v] + u.y;
                q.push({-dist[u.x], u.x});
            }
        }
    }

    if (dist[n - 1] == inf) dist[n - 1] = -1;

    cout << dist[n - 1];

    return 0;
}



```