#include <bits/stdc++.h>
using namespace std;
void solveA() {
  int n;
  cin >> n;
  vector<long double> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  sort(a.begin(), a.end());
  long double ans = -2000000000.0;
  long double sum2 = 0;
  for (int i = 0; i < n; ++i) sum2 += a[i];
  long double sum1 = 0;
  for (int i = 0; i + 1 < n; ++i) {
    sum1 += a[i];
    sum2 -= a[i];
    ans = max(ans, sum1 / (i + 1) + sum2 / (n - i - 1));
  }
  cout << ans << "\n";
}
void solveB() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  vector<int> b = a;
  sort(b.begin(), b.end());
  b.resize(unique(b.begin(), b.end()) - b.begin());
  for (int i = 0; i < n; ++i)
    a[i] = lower_bound(b.begin(), b.end(), a[i]) - b.begin();
  int kol = 1;
  for (int i = 1; i < n; ++i)
    if (!((a[i] == a[i - 1] + 1))) ++kol;
  if (kol <= k)
    cout << "YES\n";
  else
    cout << "NO\n";
}
const int MOD = 1000000007;
long long powmod(long long a, long long n) {
  long long ans = 1;
  while (n > 0) {
    if (n % 2 == 1) ans = (ans * a) % MOD;
    a = (a * a) % MOD;
    n /= 2;
  }
  return ans;
}
void solveC() {
  long long n, k;
  cin >> n >> k;
  if (k == 0) {
    cout << "1\n";
    return;
  }
  long long kolEq = powmod(2, n - 1);
  if (n % 2 == 0)
    --kolEq;
  else
    ++kolEq;
  if (kolEq < 0) kolEq += MOD;
  kolEq %= MOD;
  long long kolMore = (n % 2 == 0);
  long long z = powmod(2, n);
  z = powmod(z, k - 1);
  long long f = 1;
  for (int i = 1; i <= k; ++i) {
    f = kolEq * f + kolMore * powmod(2, n * (i - 1));
    f %= MOD;
  }
  cout << f << "\n";
}
struct Seg {
  int i, l, r;
};
bool operator<(Seg s1, Seg s2) { return s1.i < s2.i; }
struct Node {
  int mx, pos;
  bool isAss;
  int assMx, assPos;
};
vector<Node> tree;
void assignOnVertex(int v, int l, int r, pair<int, int> p) {
  tree[v].isAss = true;
  tree[v].mx = tree[v].assMx = p.first;
  tree[v].pos = tree[v].assPos = p.second;
}
void push(int v, int l, int r) {
  if (tree[v].isAss) {
    pair<int, int> p(tree[v].assMx, tree[v].assPos);
    int mid = (l + r) / 2;
    assignOnVertex(v * 2, l, mid, p);
    assignOnVertex(v * 2 + 1, mid + 1, r, p);
    tree[v].isAss = false;
  }
}
pair<int, int> findMax(int v, int l, int r, int ql, int qr) {
  if (r < ql || qr < l) return make_pair(-1, -1);
  if (ql <= l && r <= qr) return make_pair(tree[v].mx, tree[v].pos);
  push(v, l, r);
  int mid = (l + r) / 2;
  return max(findMax(v * 2, l, mid, ql, qr),
             findMax(v * 2 + 1, mid + 1, r, ql, qr));
}
pair<int, int> findMaxInTree(int l, int r) {
  return findMax(1, 0, tree.size() / 2 - 1, l, r);
}
void assign(int v, int l, int r, int ql, int qr, pair<int, int> qp) {
  if (r < ql || qr < l) return;
  if (ql <= l && r <= qr) {
    assignOnVertex(v, l, r, qp);
    return;
  }
  push(v, l, r);
  int mid = (l + r) / 2;
  assign(v * 2, l, mid, ql, qr, qp);
  assign(v * 2 + 1, mid + 1, r, ql, qr, qp);
  pair<int, int> p = max(make_pair(tree[2 * v].mx, tree[2 * v].pos),
                         make_pair(tree[2 * v + 1].mx, tree[2 * v + 1].pos));
  tree[v].mx = p.first;
  tree[v].pos = p.second;
}
void assignInTree(int l, int r, pair<int, int> pp) {
  assign(1, 0, tree.size() / 2 - 1, l, r, pp);
}
void solveD() {
  int n, m;
  cin >> n >> m;
  vector<Seg> ss(m);
  for (int i = 0; i < m; ++i) cin >> ss[i].i >> ss[i].l >> ss[i].r;
  vector<int> cs;
  for (int i = 0; i < m; ++i) {
    ss[i].i--;
    cs.push_back(ss[i].l);
    cs.push_back(ss[i].r);
  }
  sort(cs.begin(), cs.end());
  cs.resize(unique(cs.begin(), cs.end()) - cs.begin());
  for (int i = 0; i < m; ++i) {
    ss[i].l = lower_bound(cs.begin(), cs.end(), ss[i].l) - cs.begin();
    ss[i].r = lower_bound(cs.begin(), cs.end(), ss[i].r) - cs.begin();
  }
  sort(ss.begin(), ss.end());
  int deg = 1;
  while (deg < cs.size()) deg *= 2;
  tree.assign(deg * 2, Node{0, -1, false, -1, -1});
  vector<int> d(n, -1);
  vector<int> pr(n, -1);
  for (int i = 0; i < m;) {
    int ni = i;
    while (ni < m && ss[i].i == ss[ni].i) ++ni;
    pair<int, int> ans(-1, -1);
    for (int j = i; j < ni; ++j)
      ans = max(ans, findMaxInTree(ss[j].l, ss[j].r));
    d[ss[i].i] = ans.first + 1;
    pr[ss[i].i] = ans.second;
    for (int j = i; j < ni; ++j)
      assignInTree(ss[j].l, ss[j].r, make_pair(ans.first + 1, ss[i].i));
    i = ni;
  }
  int pos = max_element(d.begin(), d.end()) - d.begin();
  int ans = d[pos];
  vector<int> used(n, 0);
  for (; pos >= 0; pos = pr[pos]) used[pos] = 1;
  cout << n - ans << "\n";
  for (int i = 0; i < n; ++i)
    if (!used[i]) cout << i + 1 << ' ';
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  int test = 1;
  while (test--) solveD();
  return 0;
}

### Prompt

Your task is to create a Cpp solution to 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;
void solveA() {
  int n;
  cin >> n;
  vector<long double> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  sort(a.begin(), a.end());
  long double ans = -2000000000.0;
  long double sum2 = 0;
  for (int i = 0; i < n; ++i) sum2 += a[i];
  long double sum1 = 0;
  for (int i = 0; i + 1 < n; ++i) {
    sum1 += a[i];
    sum2 -= a[i];
    ans = max(ans, sum1 / (i + 1) + sum2 / (n - i - 1));
  }
  cout << ans << "\n";
}
void solveB() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  vector<int> b = a;
  sort(b.begin(), b.end());
  b.resize(unique(b.begin(), b.end()) - b.begin());
  for (int i = 0; i < n; ++i)
    a[i] = lower_bound(b.begin(), b.end(), a[i]) - b.begin();
  int kol = 1;
  for (int i = 1; i < n; ++i)
    if (!((a[i] == a[i - 1] + 1))) ++kol;
  if (kol <= k)
    cout << "YES\n";
  else
    cout << "NO\n";
}
const int MOD = 1000000007;
long long powmod(long long a, long long n) {
  long long ans = 1;
  while (n > 0) {
    if (n % 2 == 1) ans = (ans * a) % MOD;
    a = (a * a) % MOD;
    n /= 2;
  }
  return ans;
}
void solveC() {
  long long n, k;
  cin >> n >> k;
  if (k == 0) {
    cout << "1\n";
    return;
  }
  long long kolEq = powmod(2, n - 1);
  if (n % 2 == 0)
    --kolEq;
  else
    ++kolEq;
  if (kolEq < 0) kolEq += MOD;
  kolEq %= MOD;
  long long kolMore = (n % 2 == 0);
  long long z = powmod(2, n);
  z = powmod(z, k - 1);
  long long f = 1;
  for (int i = 1; i <= k; ++i) {
    f = kolEq * f + kolMore * powmod(2, n * (i - 1));
    f %= MOD;
  }
  cout << f << "\n";
}
struct Seg {
  int i, l, r;
};
bool operator<(Seg s1, Seg s2) { return s1.i < s2.i; }
struct Node {
  int mx, pos;
  bool isAss;
  int assMx, assPos;
};
vector<Node> tree;
void assignOnVertex(int v, int l, int r, pair<int, int> p) {
  tree[v].isAss = true;
  tree[v].mx = tree[v].assMx = p.first;
  tree[v].pos = tree[v].assPos = p.second;
}
void push(int v, int l, int r) {
  if (tree[v].isAss) {
    pair<int, int> p(tree[v].assMx, tree[v].assPos);
    int mid = (l + r) / 2;
    assignOnVertex(v * 2, l, mid, p);
    assignOnVertex(v * 2 + 1, mid + 1, r, p);
    tree[v].isAss = false;
  }
}
pair<int, int> findMax(int v, int l, int r, int ql, int qr) {
  if (r < ql || qr < l) return make_pair(-1, -1);
  if (ql <= l && r <= qr) return make_pair(tree[v].mx, tree[v].pos);
  push(v, l, r);
  int mid = (l + r) / 2;
  return max(findMax(v * 2, l, mid, ql, qr),
             findMax(v * 2 + 1, mid + 1, r, ql, qr));
}
pair<int, int> findMaxInTree(int l, int r) {
  return findMax(1, 0, tree.size() / 2 - 1, l, r);
}
void assign(int v, int l, int r, int ql, int qr, pair<int, int> qp) {
  if (r < ql || qr < l) return;
  if (ql <= l && r <= qr) {
    assignOnVertex(v, l, r, qp);
    return;
  }
  push(v, l, r);
  int mid = (l + r) / 2;
  assign(v * 2, l, mid, ql, qr, qp);
  assign(v * 2 + 1, mid + 1, r, ql, qr, qp);
  pair<int, int> p = max(make_pair(tree[2 * v].mx, tree[2 * v].pos),
                         make_pair(tree[2 * v + 1].mx, tree[2 * v + 1].pos));
  tree[v].mx = p.first;
  tree[v].pos = p.second;
}
void assignInTree(int l, int r, pair<int, int> pp) {
  assign(1, 0, tree.size() / 2 - 1, l, r, pp);
}
void solveD() {
  int n, m;
  cin >> n >> m;
  vector<Seg> ss(m);
  for (int i = 0; i < m; ++i) cin >> ss[i].i >> ss[i].l >> ss[i].r;
  vector<int> cs;
  for (int i = 0; i < m; ++i) {
    ss[i].i--;
    cs.push_back(ss[i].l);
    cs.push_back(ss[i].r);
  }
  sort(cs.begin(), cs.end());
  cs.resize(unique(cs.begin(), cs.end()) - cs.begin());
  for (int i = 0; i < m; ++i) {
    ss[i].l = lower_bound(cs.begin(), cs.end(), ss[i].l) - cs.begin();
    ss[i].r = lower_bound(cs.begin(), cs.end(), ss[i].r) - cs.begin();
  }
  sort(ss.begin(), ss.end());
  int deg = 1;
  while (deg < cs.size()) deg *= 2;
  tree.assign(deg * 2, Node{0, -1, false, -1, -1});
  vector<int> d(n, -1);
  vector<int> pr(n, -1);
  for (int i = 0; i < m;) {
    int ni = i;
    while (ni < m && ss[i].i == ss[ni].i) ++ni;
    pair<int, int> ans(-1, -1);
    for (int j = i; j < ni; ++j)
      ans = max(ans, findMaxInTree(ss[j].l, ss[j].r));
    d[ss[i].i] = ans.first + 1;
    pr[ss[i].i] = ans.second;
    for (int j = i; j < ni; ++j)
      assignInTree(ss[j].l, ss[j].r, make_pair(ans.first + 1, ss[i].i));
    i = ni;
  }
  int pos = max_element(d.begin(), d.end()) - d.begin();
  int ans = d[pos];
  vector<int> used(n, 0);
  for (; pos >= 0; pos = pr[pos]) used[pos] = 1;
  cout << n - ans << "\n";
  for (int i = 0; i < n; ++i)
    if (!used[i]) cout << i + 1 << ' ';
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  int test = 1;
  while (test--) solveD();
  return 0;
}

```