#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll getrnd(ll l, ll r) { return uniform_int_distribution<ll>(l, r)(rng); }
template <typename T1, typename T2>
inline bool relax(T1& a, const T2& b) {
  return a > b ? a = b, true : false;
}
template <typename T1, typename T2>
inline bool strain(T1& a, const T2& b) {
  return a < b ? a = b, true : false;
}
const int N = 6e5 + 5;
pair<int, int> tree[4 * N], lazy[4 * N];
pair<int, int> merge(pair<int, int> a, pair<int, int> b) {
  if (a.first > b.first) return a;
  return b;
}
void upd(int v, pair<int, int> x) {
  tree[v] = merge(tree[v], x);
  lazy[v] = merge(lazy[v], x);
}
void push(int v) {
  if (lazy[v].first >= 0) {
    upd(2 * v, lazy[v]);
    upd(2 * v + 1, lazy[v]);
    lazy[v].first = -1;
  }
}
void build(int v, int tl, int tr) {
  if (tl == tr)
    tree[v] = {-1, tl}, lazy[v] = {-1, -1};
  else {
    int tm = (tl + tr) / 2;
    build(2 * v, tl, tm);
    build(2 * v + 1, tm + 1, tr);
    tree[v] = merge(tree[2 * v], tree[2 * v + 1]);
  }
}
void update(int v, int tl, int tr, int l, int r, pair<int, int> x) {
  if (l > r) return;
  if (tl == l && tr == r)
    upd(v, x);
  else {
    int tm = (tl + tr) / 2;
    push(v);
    update(2 * v, tl, tm, l, min(r, tm), x);
    update(2 * v + 1, tm + 1, tr, max(l, tm + 1), r, x);
    tree[v] = merge(tree[2 * v], tree[2 * v + 1]);
  }
}
pair<int, int> get(int v, int tl, int tr, int l, int r) {
  if (l > r) return {-1, -1};
  if (tl == l && tr == r)
    return tree[v];
  else {
    int tm = (tl + tr) / 2;
    push(v);
    return merge(get(2 * v, tl, tm, l, min(r, tm)),
                 get(2 * v + 1, tm + 1, tr, max(l, tm + 1), r));
  }
}
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> ids;
  vector<vector<pair<int, int>>> segs(n);
  for (int i = 0; i < m; ++i) {
    int ind, l, r;
    cin >> ind >> l >> r;
    --ind;
    ids.emplace_back(l);
    ids.emplace_back(r);
    segs[ind].emplace_back(l, r);
  }
  sort(ids.begin(), ids.end());
  ids.resize(unique(ids.begin(), ids.end()) - ids.begin());
  auto gid = [&](int x) {
    return lower_bound(ids.begin(), ids.end(), x) - ids.begin();
  };
  int sz = ids.size();
  build(1, 0, sz);
  vector<int> dp(n, 1), from(n, -1);
  for (int i = 0; i < n; ++i) {
    for (auto& [l, r] : segs[i]) {
      l = gid(l);
      r = gid(r);
      auto [val, ind] = get(1, 0, sz, l, r);
      if (strain(dp[i], val + 1)) from[i] = ind;
    }
    for (auto [l, r] : segs[i]) {
      update(1, 0, sz, l, r, {dp[i], i});
    }
  }
  int st = max_element(dp.begin(), dp.end()) - dp.begin();
  cout << n - dp[st] << '\n';
  vector<int> was(n);
  while (st != -1) {
    was[st] = 1;
    st = from[st];
  }
  for (int i = 0; i < n; ++i)
    if (!was[i]) cout << i + 1 << ' ';
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#pragma GCC optimize("Ofast")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll getrnd(ll l, ll r) { return uniform_int_distribution<ll>(l, r)(rng); }
template <typename T1, typename T2>
inline bool relax(T1& a, const T2& b) {
  return a > b ? a = b, true : false;
}
template <typename T1, typename T2>
inline bool strain(T1& a, const T2& b) {
  return a < b ? a = b, true : false;
}
const int N = 6e5 + 5;
pair<int, int> tree[4 * N], lazy[4 * N];
pair<int, int> merge(pair<int, int> a, pair<int, int> b) {
  if (a.first > b.first) return a;
  return b;
}
void upd(int v, pair<int, int> x) {
  tree[v] = merge(tree[v], x);
  lazy[v] = merge(lazy[v], x);
}
void push(int v) {
  if (lazy[v].first >= 0) {
    upd(2 * v, lazy[v]);
    upd(2 * v + 1, lazy[v]);
    lazy[v].first = -1;
  }
}
void build(int v, int tl, int tr) {
  if (tl == tr)
    tree[v] = {-1, tl}, lazy[v] = {-1, -1};
  else {
    int tm = (tl + tr) / 2;
    build(2 * v, tl, tm);
    build(2 * v + 1, tm + 1, tr);
    tree[v] = merge(tree[2 * v], tree[2 * v + 1]);
  }
}
void update(int v, int tl, int tr, int l, int r, pair<int, int> x) {
  if (l > r) return;
  if (tl == l && tr == r)
    upd(v, x);
  else {
    int tm = (tl + tr) / 2;
    push(v);
    update(2 * v, tl, tm, l, min(r, tm), x);
    update(2 * v + 1, tm + 1, tr, max(l, tm + 1), r, x);
    tree[v] = merge(tree[2 * v], tree[2 * v + 1]);
  }
}
pair<int, int> get(int v, int tl, int tr, int l, int r) {
  if (l > r) return {-1, -1};
  if (tl == l && tr == r)
    return tree[v];
  else {
    int tm = (tl + tr) / 2;
    push(v);
    return merge(get(2 * v, tl, tm, l, min(r, tm)),
                 get(2 * v + 1, tm + 1, tr, max(l, tm + 1), r));
  }
}
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> ids;
  vector<vector<pair<int, int>>> segs(n);
  for (int i = 0; i < m; ++i) {
    int ind, l, r;
    cin >> ind >> l >> r;
    --ind;
    ids.emplace_back(l);
    ids.emplace_back(r);
    segs[ind].emplace_back(l, r);
  }
  sort(ids.begin(), ids.end());
  ids.resize(unique(ids.begin(), ids.end()) - ids.begin());
  auto gid = [&](int x) {
    return lower_bound(ids.begin(), ids.end(), x) - ids.begin();
  };
  int sz = ids.size();
  build(1, 0, sz);
  vector<int> dp(n, 1), from(n, -1);
  for (int i = 0; i < n; ++i) {
    for (auto& [l, r] : segs[i]) {
      l = gid(l);
      r = gid(r);
      auto [val, ind] = get(1, 0, sz, l, r);
      if (strain(dp[i], val + 1)) from[i] = ind;
    }
    for (auto [l, r] : segs[i]) {
      update(1, 0, sz, l, r, {dp[i], i});
    }
  }
  int st = max_element(dp.begin(), dp.end()) - dp.begin();
  cout << n - dp[st] << '\n';
  vector<int> was(n);
  while (st != -1) {
    was[st] = 1;
    st = from[st];
  }
  for (int i = 0; i < n; ++i)
    if (!was[i]) cout << i + 1 << ' ';
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t = 1;
  while (t--) solve();
  return 0;
}

```