#include <bits/stdc++.h>
using namespace std;

const int N = 100010;
const int MAXA = 100001;          // a 的最大值（读入后 +1）
const int MAXB = 200000;          // b 的最大值（读入后 +1）

int n, m, x[N], a[N], b[N], t1[MAXB + 5], t2[MAXB + 5];
vector<int> idx[MAXA + 5];

inline int lowbit(int x) { return x & -x; }

inline void add(int bit[], int pos, int val) {
    // 统一用 MAXB 作为 Fenwick 的上界
    for (int i = pos; i <= MAXB; i += lowbit(i)) bit[i] += val;
}

inline int sumFenwick(int bit[], int pos) {
    if (pos <= 0) return 0;        // 防止负数导致死循环
    if (pos > MAXB) pos = MAXB;    // 夹到上界
    int res = 0;
    for (int i = pos; i > 0; i -= lowbit(i)) res += bit[i];
    return res;
}

int main() {
    freopen("difficulty.in", "r", stdin);
    freopen("difficulty.out", "w", stdout);
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        scanf("%d%d%d", &x[i], &a[i], &b[i]);
        a[i]++; b[i]++;
        idx[a[i]].push_back(i);
    }
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
        if (x[i] == 3) cnt++;
        else if (x[i] == 2) add(t2, b[i], 1);
        else if (x[i] == 1) add(t1, b[i], 1);
    }
    int ans = 1e9;

    for (int i = MAXA; i >= 0; i--) {
        for (int w : idx[i]) {
            if (x[w] == 2) {
                cnt++;
                add(t2, b[w], -1);
            } else if (x[w] == 1) {
                add(t1, b[w], -1);
                add(t2, b[w], 1);
            } else if (x[w] == 0) {
                add(t1, b[w], 1);
            }
        }

        int l = 0, r = MAXB;
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            int tmp = cnt + sumFenwick(t2, MAXB) + sumFenwick(t1, MAXB) - sumFenwick(t1, mid - 1);
            if (tmp >= m) l = mid;
            else r = mid - 1;
        }
        if (l == 0) continue;
        int tmp = cnt + sumFenwick(t2, MAXB) - sumFenwick(t2, l - 1);
        ans = min(ans, tmp);
    }
    printf("%d\n", ans);
    return 0;
}