#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
struct SegTree {
  vector<pii> st;
  vector<pii> lazy;
  SegTree(int n) : st(4 * n + 1), lazy(4 * n + 1){};
  void push(int v) {
    for (int ch = v * 2; ch <= v * 2 + 1; ch++) {
      st[ch] = max(st[ch], lazy[v]);
      lazy[ch] = max(lazy[ch], lazy[v]);
    }
    lazy[v] = {0, 0};
  }
  void update(int v, int tl, int tr, int l, int r, pii val) {
    if (l > r) return;
    if (tl == l && tr == r) {
      st[v] = max(val, st[v]);
      lazy[v] = max(val, lazy[v]);
      return;
    }
    push(v);
    int tm = (tl + tr) / 2;
    update(v * 2, tl, tm, l, min(r, tm), val);
    update(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, val);
    st[v] = max(st[v * 2], st[v * 2 + 1]);
  }
  pii query(int v, int tl, int tr, int l, int r) {
    if (l > r) return {0, 0};
    if (tl == l && tr == r) return st[v];
    push(v);
    int tm = (tl + tr) / 2;
    pii ql = query(v * 2, tl, tm, l, min(r, tm));
    pii qr = query(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
    return max(ql, qr);
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  map<int, int> comp;
  vector<vector<pii>> seg(n + 1);
  for (int j = 0; j < m; j++) {
    int i, l, r;
    cin >> i >> l >> r;
    seg[i].push_back({l, r});
    comp[l] = 0;
    comp[r] = 0;
  }
  int num = 0;
  for (auto &[k, v] : comp) {
    v = ++num;
  }
  SegTree st(num);
  vector<int> to(n + 5);
  for (int i = 1; i <= n; i++) {
    pii mx = {0, 0};
    for (auto &[l, r] : seg[i]) {
      l = comp[l], r = comp[r];
      mx = max(mx, st.query(1, 1, num, l, r));
    }
    to[i] = mx.second;
    mx = {mx.first + 1, i};
    for (auto [l, r] : seg[i]) {
      st.update(1, 1, num, l, r, mx);
    }
  }
  auto [ans, r] = st.st[1];
  to[n + 1] = r;
  cout << n - ans << '\n';
  int rw = n + 1;
  while (rw) {
    for (int i = to[rw] + 1; i < rw; i++) cout << i << ' ';
    rw = to[rw];
  }
  cout << '\n';
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
struct SegTree {
  vector<pii> st;
  vector<pii> lazy;
  SegTree(int n) : st(4 * n + 1), lazy(4 * n + 1){};
  void push(int v) {
    for (int ch = v * 2; ch <= v * 2 + 1; ch++) {
      st[ch] = max(st[ch], lazy[v]);
      lazy[ch] = max(lazy[ch], lazy[v]);
    }
    lazy[v] = {0, 0};
  }
  void update(int v, int tl, int tr, int l, int r, pii val) {
    if (l > r) return;
    if (tl == l && tr == r) {
      st[v] = max(val, st[v]);
      lazy[v] = max(val, lazy[v]);
      return;
    }
    push(v);
    int tm = (tl + tr) / 2;
    update(v * 2, tl, tm, l, min(r, tm), val);
    update(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, val);
    st[v] = max(st[v * 2], st[v * 2 + 1]);
  }
  pii query(int v, int tl, int tr, int l, int r) {
    if (l > r) return {0, 0};
    if (tl == l && tr == r) return st[v];
    push(v);
    int tm = (tl + tr) / 2;
    pii ql = query(v * 2, tl, tm, l, min(r, tm));
    pii qr = query(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
    return max(ql, qr);
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  map<int, int> comp;
  vector<vector<pii>> seg(n + 1);
  for (int j = 0; j < m; j++) {
    int i, l, r;
    cin >> i >> l >> r;
    seg[i].push_back({l, r});
    comp[l] = 0;
    comp[r] = 0;
  }
  int num = 0;
  for (auto &[k, v] : comp) {
    v = ++num;
  }
  SegTree st(num);
  vector<int> to(n + 5);
  for (int i = 1; i <= n; i++) {
    pii mx = {0, 0};
    for (auto &[l, r] : seg[i]) {
      l = comp[l], r = comp[r];
      mx = max(mx, st.query(1, 1, num, l, r));
    }
    to[i] = mx.second;
    mx = {mx.first + 1, i};
    for (auto [l, r] : seg[i]) {
      st.update(1, 1, num, l, r, mx);
    }
  }
  auto [ans, r] = st.st[1];
  to[n + 1] = r;
  cout << n - ans << '\n';
  int rw = n + 1;
  while (rw) {
    for (int i = to[rw] + 1; i < rw; i++) cout << i << ' ';
    rw = to[rw];
  }
  cout << '\n';
  return 0;
}

```