#include <bits/stdc++.h>
using namespace std;
const int maxn = 300005;
int n, m, need[maxn];
vector<pair<int, int> > row[maxn];
int dp[maxn], tr[maxn];
pair<int, int> tree[maxn * 8];
pair<int, int> lazy[maxn * 8];
void PushUp(int node) {
  if (tree[node * 2] > tree[node * 2 + 1])
    tree[node] = tree[node * 2];
  else
    tree[node] = tree[node * 2 + 1];
}
void PushDown(int node, int l, int r) {
  if (tree[node].first < lazy[node].first) tree[node] = lazy[node];
  if (l == r) return;
  if (lazy[node * 2].first < lazy[node].first) {
    lazy[node * 2] = lazy[node];
  }
  if (lazy[node * 2 + 1].first < lazy[node].first) {
    lazy[node * 2 + 1] = lazy[node];
  }
  lazy[node] = {0, 0};
}
void Update(int node, int l, int r, int u, int v, int id, int val) {
  PushDown(node, l, r);
  if (l > v || r < u) return;
  if (l >= u && r <= v) {
    if (tree[node].first < val) {
      tree[node] = {val, id};
    }
    if (lazy[node].first < val) {
      lazy[node] = {val, id};
    }
    return;
  }
  int mid = (l + r) >> 1;
  Update(node * 2, l, mid, u, v, id, val);
  Update(node * 2 + 1, mid + 1, r, u, v, id, val);
  PushUp(node);
}
pair<int, int> Get(int node, int l, int r, int u, int v) {
  PushDown(node, l, r);
  if (l > v || r < u) return {0, 0};
  if (l >= u && r <= v) return tree[node];
  int mid = (l + r) >> 1;
  pair<int, int> L = Get(node * 2, l, mid, u, v);
  pair<int, int> R = Get(node * 2 + 1, mid + 1, r, u, v);
  if (L.first > R.first) return L;
  return R;
}
void Solve(int Test) {
  cin >> n >> m;
  vector<int> v;
  for (int i = 1; i <= m; ++i) {
    int id, l, r;
    cin >> id >> l >> r;
    v.push_back(l), v.push_back(r);
    row[id].push_back({l, r});
  }
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  m = v.size();
  for (int i = 1; i <= n; ++i) {
    for (auto &p : row[i]) {
      int &l = p.first, &r = p.second;
      l = lower_bound(v.begin(), v.end(), l) - v.begin() + 1;
      r = lower_bound(v.begin(), v.end(), r) - v.begin() + 1;
      pair<int, int> cur = Get(1, 1, m, l, r);
      if (dp[i] < cur.first + 1) {
        dp[i] = cur.first + 1;
        tr[i] = cur.second;
      }
    }
    for (auto p : row[i]) {
      int l = p.first, r = p.second;
      Update(1, 1, m, l, r, i, dp[i]);
    }
  }
  int pos = 0;
  for (int i = 1; i <= n; ++i) {
    if (dp[i] > dp[pos]) {
      pos = i;
    }
  }
  int res = 0;
  while (pos) {
    need[pos] = 1;
    pos = tr[pos];
    res++;
  }
  cout << n - res << "\n";
  for (int i = 1; i <= n; ++i) {
    if (!need[i]) cout << i << " ";
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  if (fopen("CF"
            ".inp",
            "r")) {
    freopen(
        "CF"
        ".inp",
        "r", stdin);
    freopen(
        "CF"
        ".out",
        "w", stdout);
  }
  int test = 1;
  for (int i = 1; i <= test; ++i) {
    Solve(i);
  }
}

### Prompt

Your challenge is to write 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;
const int maxn = 300005;
int n, m, need[maxn];
vector<pair<int, int> > row[maxn];
int dp[maxn], tr[maxn];
pair<int, int> tree[maxn * 8];
pair<int, int> lazy[maxn * 8];
void PushUp(int node) {
  if (tree[node * 2] > tree[node * 2 + 1])
    tree[node] = tree[node * 2];
  else
    tree[node] = tree[node * 2 + 1];
}
void PushDown(int node, int l, int r) {
  if (tree[node].first < lazy[node].first) tree[node] = lazy[node];
  if (l == r) return;
  if (lazy[node * 2].first < lazy[node].first) {
    lazy[node * 2] = lazy[node];
  }
  if (lazy[node * 2 + 1].first < lazy[node].first) {
    lazy[node * 2 + 1] = lazy[node];
  }
  lazy[node] = {0, 0};
}
void Update(int node, int l, int r, int u, int v, int id, int val) {
  PushDown(node, l, r);
  if (l > v || r < u) return;
  if (l >= u && r <= v) {
    if (tree[node].first < val) {
      tree[node] = {val, id};
    }
    if (lazy[node].first < val) {
      lazy[node] = {val, id};
    }
    return;
  }
  int mid = (l + r) >> 1;
  Update(node * 2, l, mid, u, v, id, val);
  Update(node * 2 + 1, mid + 1, r, u, v, id, val);
  PushUp(node);
}
pair<int, int> Get(int node, int l, int r, int u, int v) {
  PushDown(node, l, r);
  if (l > v || r < u) return {0, 0};
  if (l >= u && r <= v) return tree[node];
  int mid = (l + r) >> 1;
  pair<int, int> L = Get(node * 2, l, mid, u, v);
  pair<int, int> R = Get(node * 2 + 1, mid + 1, r, u, v);
  if (L.first > R.first) return L;
  return R;
}
void Solve(int Test) {
  cin >> n >> m;
  vector<int> v;
  for (int i = 1; i <= m; ++i) {
    int id, l, r;
    cin >> id >> l >> r;
    v.push_back(l), v.push_back(r);
    row[id].push_back({l, r});
  }
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  m = v.size();
  for (int i = 1; i <= n; ++i) {
    for (auto &p : row[i]) {
      int &l = p.first, &r = p.second;
      l = lower_bound(v.begin(), v.end(), l) - v.begin() + 1;
      r = lower_bound(v.begin(), v.end(), r) - v.begin() + 1;
      pair<int, int> cur = Get(1, 1, m, l, r);
      if (dp[i] < cur.first + 1) {
        dp[i] = cur.first + 1;
        tr[i] = cur.second;
      }
    }
    for (auto p : row[i]) {
      int l = p.first, r = p.second;
      Update(1, 1, m, l, r, i, dp[i]);
    }
  }
  int pos = 0;
  for (int i = 1; i <= n; ++i) {
    if (dp[i] > dp[pos]) {
      pos = i;
    }
  }
  int res = 0;
  while (pos) {
    need[pos] = 1;
    pos = tr[pos];
    res++;
  }
  cout << n - res << "\n";
  for (int i = 1; i <= n; ++i) {
    if (!need[i]) cout << i << " ";
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  if (fopen("CF"
            ".inp",
            "r")) {
    freopen(
        "CF"
        ".inp",
        "r", stdin);
    freopen(
        "CF"
        ".out",
        "w", stdout);
  }
  int test = 1;
  for (int i = 1; i <= test; ++i) {
    Solve(i);
  }
}

```