description: Moamen was drawing a grid of n rows and 10^9 columns containing only digits 0 and 1. Ezzat noticed what Moamen was drawing and became interested in the minimum number of rows one needs to remove to make the grid beautiful.
  
  A grid is beautiful if and only if for every two consecutive rows there is at least one column containing 1 in these two rows.
  
  Ezzat will give you the number of rows n, and m segments of the grid that contain digits 1. Every segment is represented with three integers i, l, and r, where i represents the row number, and l and r represent the first and the last column of the segment in that row.
  
  For example, if n = 3, m = 6, and the segments are (1,1,1), (1,7,8), (2,7,7), (2,15,15), (3,1,1), (3,15,15), then the grid is:
  
  <image>
  
  Your task is to tell Ezzat the minimum number of rows that should be removed to make the grid beautiful.
  
  Input
  
  The first line contains two integers n and m (1 ≤ n, m ≤ 3⋅10^5).
  
  Each of the next m lines contains three integers i, l, and r (1 ≤ i ≤ n, 1 ≤ l ≤ r ≤ 10^9). Each of these m lines means that row number i contains digits 1 in columns from l to r, inclusive.
  
  Note that the segments may overlap.
  
  Output
  
  In the first line, print a single integer k — the minimum number of rows that should be removed.
  
  In the second line print k distinct integers r_1, r_2, …, r_k, representing the rows that should be removed (1 ≤ r_i ≤ n), in any order.
  
  If there are multiple answers, print any.
  
  Examples
  
  Input
  
  
  3 6
  1 1 1
  1 7 8
  2 7 7
  2 15 15
  3 1 1
  3 15 15
  
  
  Output
  
  
  0
  
  
  Input
  
  
  5 4
  1 2 3
  2 4 6
  3 3 5
  5 1 1
  
  
  Output
  
  
  3
  2 4 5
  
  Note
  
  In the first test case, the grid is the one explained in the problem statement. The grid has the following properties: 
  
    1. The 1-st row and the 2-nd row have a common 1 in the column 7. 
    2. The 2-nd row and the 3-rd row have a common 1 in the column 15. 
  
  As a result, this grid is beautiful and we do not need to remove any row.
  
  In the second test case, the given grid is as follows: 
  
  <image>
solution: #include <bits/stdc++.h>
  using namespace std;
  const int maxn = 3e5 + 10;
  int n, m, ls[maxn * 3], pos, dp[maxn], last[maxn];
  bool vis[maxn];
  int madoka[maxn];
  struct Node {
    int l, r;
    Node() {}
    Node(int l, int r) : l(l), r(r) {}
  };
  struct node {
    int l, r, id;
    int val;
  } fbk[maxn << 4];
  struct LLL {
    int id;
    int w;
  } lazy[maxn << 4];
  void pushup(int x) {
    if (fbk[x << 1].val >= fbk[x << 1 | 1].val)
      fbk[x].id = fbk[x << 1].id;
    else
      fbk[x].id = fbk[x << 1 | 1].id;
    fbk[x].val = max(fbk[x << 1].val, fbk[x << 1 | 1].val);
    return;
  }
  void pushdown(int x) {
    if (lazy[x].w) {
      int w = lazy[x].w;
      int id = lazy[x].id;
      lazy[x].w = lazy[x].id = 0;
      if (w > lazy[x << 1].w) {
        lazy[x << 1].w = w;
        lazy[x << 1].id = id;
      }
      if (w > lazy[x << 1 | 1].w) {
        lazy[x << 1 | 1].w = w;
        lazy[x << 1 | 1].id = id;
      }
      if (w > fbk[x << 1].val) {
        fbk[x << 1].val = w;
        fbk[x << 1].id = id;
      }
      if (w > fbk[x << 1 | 1].val) {
        fbk[x << 1 | 1].val = w;
        fbk[x << 1 | 1].id = id;
      }
    }
    return;
  }
  void build(int l, int r, int x) {
    fbk[x].l = l;
    fbk[x].r = r;
    fbk[x].val = fbk[x].id = 0;
    lazy[x].id = lazy[x].w = 0;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(l, mid, x << 1);
    build(mid + 1, r, x << 1 | 1);
    return;
  }
  node query(int L, int R, int x) {
    int l = fbk[x].l;
    int r = fbk[x].r;
    if (l >= L && r <= R) return fbk[x];
    pushdown(x);
    int mid = (l + r) >> 1;
    if (L > mid)
      return query(L, R, x << 1 | 1);
    else if (R <= mid)
      return query(L, R, x << 1);
    else {
      node a = query(L, mid, x << 1);
      node b = query(mid + 1, R, x << 1 | 1);
      if (a.val > b.val)
        return a;
      else
        return b;
    }
  }
  void update(int L, int R, int x, int Max, int idx) {
    int l = fbk[x].l;
    int r = fbk[x].r;
    if (l >= L && r <= R) {
      if (Max > lazy[x].w) {
        lazy[x].w = Max;
        lazy[x].id = idx;
      }
      if (Max > fbk[x].val) {
        fbk[x].val = Max;
        fbk[x].id = idx;
      }
      return;
    }
    pushdown(x);
    int mid = (l + r) >> 1;
    if (L > mid)
      update(L, R, x << 1 | 1, Max, idx);
    else if (R <= mid)
      update(L, R, x << 1, Max, idx);
    else {
      update(L, mid, x << 1, Max, idx);
      update(mid + 1, R, x << 1 | 1, Max, idx);
    }
    pushup(x);
    return;
  }
  vector<Node> E[maxn];
  int main() {
    cin >> n >> m;
    int i, j, k, x, l, r;
    for (i = 1; i <= m; i++) {
      scanf("%d%d%d", &x, &l, &r);
      E[x].push_back(Node(l, r));
      ls[++pos] = l;
      ls[++pos] = r;
    }
    sort(ls + 1, ls + pos + 1);
    pos = unique(ls + 1, ls + pos + 1) - ls - 1;
    for (i = 1; i <= n; i++) {
      for (j = 0; j < E[i].size(); j++) {
        l = E[i][j].l;
        r = E[i][j].r;
        E[i][j].l = lower_bound(ls + 1, ls + pos + 1, l) - ls;
        E[i][j].r = lower_bound(ls + 1, ls + pos + 1, r) - ls;
      }
    }
    int ans = 0;
    build(1, pos, 1);
    for (i = 1; i <= n; i++) {
      node t;
      t.id = 0;
      t.val = 1;
      for (j = 0; j < E[i].size(); j++) {
        l = E[i][j].l;
        r = E[i][j].r;
        node tt = query(l, r, 1);
        if (t.val < tt.val + 1) {
          t = tt;
          t.val++;
        }
      }
      dp[i] = t.val;
      last[i] = t.id;
      for (j = 0; j < E[i].size(); j++) {
        l = E[i][j].l;
        r = E[i][j].r;
        update(l, r, 1, dp[i], i);
      }
      ans = max(ans, dp[i]);
    }
    int root = 0;
    for (i = 1; i <= n; i++) {
      if (ans == dp[i]) {
        root = i;
        break;
      }
    }
    while (root != 0) {
      vis[root] = 1;
      root = last[root];
    }
    int len = 0;
    for (i = 1; i <= n; i++) {
      if (!vis[i]) madoka[++len] = i;
    }
    cout << len << endl;
    for (i = 1; i <= len; i++) printf(i == len ? "%d\n" : "%d ", madoka[i]);
    return 0;
  }
  