/*
①s[i] >= s[i - 1]
②s[i] - s[i - 1] <= 1 -> s[i - 1] >= s[i] - 1
③在[a , b]中存在c个 -> s[b] - s[a - 1] >= c -> s[b] >= s[a - 1] + c
由①②可以说明每一个数要么选要么不选(选是1,不选是0)
由③可以说明在区间内至少存在c个
于是满足了题目的条件!
*/
#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
const int N = 50010 , M = 3 * N;
int e[M] , ne[M] , w[M] , h[N] , idx;
int dist[N];
bool st[N];
int n;
void add(int a , int b , int c)
{
e[idx] = b , ne[idx] = h[a] , w[idx] = c , h[a] = idx++;
}
void spfa()
{
memset(dist , -0x3f , sizeof dist);
dist[0] = 0;
queue<int> q;
q.push(0);
st[0] = true;
while(q.size())
{
int t = q.front();
q.pop();
st[t] = false;
for( int i = h[t] ; ~i ; i = ne[i])
{
int j = e[i];
if(dist[j] < dist[t] + w[i])
{
dist[j] = dist[t] + w[i];
if(!st[j])
{
st[j] = true;
q.push(j);
}
}
}
}
}
int main()
{
cin >> n;
memset(h , -1 , sizeof h);
for(int i = 1 ; i <= 50001 ; i++) add(i - 1 , i , 0) , add(i , i - 1 , -1);
while(n--)
{
int a , b , c;
cin >> a >> b >> c;
a++ , b++;//因为要用到前缀和,所以整体向右移动一位。
add(a - 1 , b , c);
}
spfa();
cout << dist[50001] << endl;
return 0;
}