/*
A milking problem, the milking people have m working time periods, each time period has a certain output
One time period must be completed before work can be done another time period
After working for a period of time, must rest R time
So, to solve the problem, you can first add the end time of each time period to the R
The maximum output that can be produced is now required for a given m time period
Dp[i], which represents the maximum amount of milking produced during the I time period,
Then Dp[j] (i>j) Dp[i]=max (DP[I],DP[J]+E[I].EF)
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace Std;
struct node{int st,ed,ef;};
Node e[1005];
int n,m,r;
int dp[1005];
BOOL CMP (node A,node b) {
return a.st<b.st| | (A.st==b.st&&a.ed<b.ed);
}
int main () {
while (~SCANF ("%d%d%d", &n,&m,&r)) {
for (int i=0;i<m;i++) {
scanf ("%d%d%d", &e[i].st,&e[i].ed,&e[i].ef);
E[i].ed+=r;
}
Sort (e,e+m,cmp);
for (int i=0;i<m;i++) {
Dp[i]=e[i].ef;
for (int j=0;j<i;j++) {
if (e[j].ed<=e[i].st) {
Dp[i]=max (DP[I],DP[J]+E[I].EF);
}
}
}
int maxx=0;
for (int i=0;i<m;i++) {
Maxx=max (Maxx,dp[i]);
}
printf ("%d\n", Maxx);
}
return 0;
}
