//#pragma GCC optimize ("-O3","unroll-loops")
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<math.h>
#include<iomanip>
#include<set>
#include<numeric>
#include<cstring>
#include<cstdio>
#include<functional>
#include<bitset>
#include<limits.h>
#include<cassert>
#include<iterator>
#include<complex>


#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define REVERSE(v,n) reverse(v,v+n);
#define VREVERSE(v) reverse(v.begin(), v.end());
#define ll long long
#define pb(a) push_back(a)
#define m0(x) memset(x,0,sizeof(x))
#define print(x) cout<<x<<'\n';
#define pe(x) cout<<x<<" ";
#define lb(v,n) lower_bound(v.begin(), v.end(), n);
#define ub(v,n) upper_bound(v.begin(), v.end(), n);
#define int long long
#define all(x) (x).begin(), (x).end()
#define double long double

using namespace std;

int MOD = (ll)1000000000 + 7;
const ll INF = 1e17;
const double pi = acos(-1);
const double EPS = 1e-10;
typedef pair<int, int>P;
const int MAX = 200020;

struct sec {
	int l, p;
};
//bool cmp(const sec&s1, const sec& s2) {
//	if (s1.r == s2.r)return s1.l < s2.l;
//	else return s1.r < s2.r;
//}
vector<sec>V[100010];

int dp[100010][2];
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int N, M, A;
	cin >> N >> M >> A;
	REP(i, M) {
		int l, r, p; cin >> l >> r >> p;
		V[r].push_back({ l,p });
	}
	//sort(all(V), cmp);
	REP(i, N) {
		dp[i][1] = -A;
	}
	dp[0][1] = 0;
	FOR(i,1, N+1) {
		dp[i][0] = max({ dp[i][0], dp[i - 1][0] ,dp[i-1][1]});
		dp[i][1] = max({ dp[i][1], dp[i - 1][0] - A, dp[i - 1][1] - A });
		for (auto x : V[i]) {
			int l = x.l;
			int p = x.p;
			if (i == N) {
				dp[i][1] = max({ dp[i][1],dp[l-1][1] + p,dp[l-1][0] -A + p });
			}
			else dp[i][1] = max({ dp[i][1],dp[l-1][1] - A + p,dp[l-1][0] - 2 * A + p });
		}
	}
	int ans = max(dp[N][0], dp[N][1]);
	print(ans);
}