using P = pair<int, ll>;
using VP = vector<P>;

int @n, @m; ll @a;
vector<VP> e(n + 1);
rep(m) {
	int @l, @r; ll @x;
	e[r].push_back({l, x}); 
} 

VLL f(n + 1, -ll_inf); // f_{i, j} := 到第 i 个位置，最后一刀在 (j, j + 1) 之间的最大贡献
f[0] = 0;
ll mx = 0;
rep(i, 1, n + 1) {
	f[i] = mx - a;
	for(auto x : e[i]) {
		f[i] >?= f[x.first - 1] + x.second - (i == n ? 0 : a);
	}
	mx >?= f[i];
}
for(auto x : f) mx >?= x;
wt(mx);