#pragma g++ optimize("Ofast")
#pragma loop_opt(on)
#include <bits/stdc++.h>
#define pb emplace_back
#define ff first
#define ss second
#define mem(v,x) memset((v),(x),sizeof(v))
#define debug(x) cerr<<#x<<" = "<<x<<'\n'

using namespace std;
typedef int64_t ll;
constexpr int N = 500025, inf = 1e9, MOD = 1000000007;

#define int ll
struct Segtree {
    ll mx[N<<1],lz[N];
    int n;
    void init(auto &u, int _n) {
        n = _n;
        for(int i = 0; i < n; i++) lz[i] = 0;
        for(int i = 0; i < n; i++) mx[i+n] = u[i];
        for(int i = n-1; i > 0; i--) mx[i] = max(mx[i<<1], mx[i<<1|1]);
    }
    void upd(int p, ll d) {
        mx[p] += d;
        if(p < n) lz[p] += d;
    }
    void pull(int p) {
        for(; p>1; p>>=1)
            mx[p>>1] = max(mx[p],mx[p^1]) + lz[p>>1];
    }
    void push(int p) {
        for(int h = __lg(n); h >= 0; h--) {
            int i = p>>h;
            upd(i, lz[i>>1]);
            upd(i^1, lz[i>>1]);
            lz[i>>1] = 0;
        }
    }
    void add(int l, int r, ll d) {
        int L = l, R = r;
        for(l+=n, r+=n; l<r; l>>=1, r>>=1) {
            if(l&1) upd(l++, d);
            if(r&1) upd(--r, d);
        }
        pull(L+n), pull(R-1+n);
    }
} sgt;
signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    int n, d;
    cin >> n >> d;
    vector<tuple<int,int,int>> v(n);
    vector<int> u;
    for(auto &[x, a, b]: v) cin >> x >> a >> b, u.pb(x);
    u.pb(0), u.pb(d); // 要考慮位置0跟位置d
    sort(u.begin(), u.end()), u.erase(unique(u.begin(), u.end()), u.end());
    for(auto &[x, a, b]: v) x = lower_bound(u.begin(), u.end(), x) - u.begin();
    sgt.init(u, u.size());
    sort(v.begin(), v.end(), [](auto a, auto b){return get<2>(a) > get<2>(b);});
    ll ans = d;
    for(auto [x, a, b]: v) {
        sgt.add(x+1, u.size(), -a); //左閉右開, 0 base
        //debug(a);
        ll f = sgt.mx[1];
        if(f <= b)
            ans = min(ans, f);
    }
    cout << ans << '\n';
}

/*
set all i s.t. B[i] >= B active
for all i, F + sum(A[j] | X[j] < X[i] and active[j]) >= X[i]
F = max(X[i] - sum(A[j] | X[j] < X[i] and active[j]))
check if F <= B
*/