#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1000000007;
const int mxx = 300005;
vector<pair<int, int> > v[2 * mxx];
unordered_map<int, int> mp;
long long int Max[mxx];
long long int tree[8 * mxx][2];
void update(int l, int r, int L, int R, int n, long long int val) {
  if (l > R || r < L)
    return;
  else if (l >= L and r <= R) {
    tree[n][0] = val + 1ll;
    tree[n][1] = val + 1;
    return;
  }
  update(l, (l + r) / 2, L, R, n * 2, val);
  update((l + r) / 2 + 1, r, L, R, n * 2 + 1, val);
  tree[n][0] = max(tree[n * 2][0], tree[n * 2 + 1][0]);
}
long long int go(int l, int r, int L, int R, int n, long long int val) {
  if (l > R || r < L)
    return 0;
  else if (l >= L and r <= R) {
    return tree[n][0];
  }
  val = max(val, tree[n][1]);
  long long int x = max(go(l, (l + r) / 2, L, R, n * 2, val),
                        go((l + r) / 2 + 1, r, L, R, n * 2 + 1, val));
  return max(val, x);
}
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> compress, del;
  for (int i = 1; i <= m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    v[a].push_back({b, c});
    compress.push_back(b);
    compress.push_back(c);
  }
  sort(compress.begin(), compress.end());
  int sum = 0;
  for (auto it : compress) {
    if (mp[it] == 0) mp[it] = ++sum;
  }
  long long int ans = 0;
  int p = 0;
  for (int i = n; i >= 1; i--) {
    long long int mx = 0;
    for (auto it : v[i]) {
      int x = mp[it.first];
      int y = mp[it.second];
      long long int fre = go(1, sum, x, y, 1, 0);
      mx = max(mx, fre);
    }
    for (auto it : v[i]) {
      int x = mp[it.first];
      int y = mp[it.second];
      update(1, sum, x, y, 1, mx);
    }
    Max[i] = mx + 1;
    if (ans < mx + 1) {
      ans = mx + 1;
      p = i;
    }
  }
  memset(tree, 0, sizeof(tree));
  for (int i = 1; i <= n; i++) {
    if (i < p) {
      del.push_back(i);
    } else if (i == p) {
      for (auto it : v[i]) {
        update(1, sum, mp[it.first], mp[it.second], 1, 1);
      }
    } else if (i > p) {
      if (Max[i] < ans - 1) {
        del.push_back(i);
        continue;
      }
      int mx = 0;
      for (auto it : v[i]) {
        int x = mp[it.first];
        int y = mp[it.second];
        int fre = go(1, sum, x, y, 1, 0);
        mx = max(mx, fre);
      }
      if (mx) {
        for (auto it : v[p]) {
          update(1, sum, mp[it.first], mp[it.second], 1, -1);
        }
        ans--;
        for (auto it : v[i]) {
          update(1, sum, mp[it.first], mp[it.second], 1, 1);
        }
        p = i;
      } else
        del.push_back(i);
    }
  }
  cout << del.size() << "\n";
  for (auto it : del) {
    cout << it << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  int t = 1;
  while (t--) solve();
}

### Prompt

Develop a solution in CPP to the problem described below:

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 long long int mod = 1000000007;
const int mxx = 300005;
vector<pair<int, int> > v[2 * mxx];
unordered_map<int, int> mp;
long long int Max[mxx];
long long int tree[8 * mxx][2];
void update(int l, int r, int L, int R, int n, long long int val) {
  if (l > R || r < L)
    return;
  else if (l >= L and r <= R) {
    tree[n][0] = val + 1ll;
    tree[n][1] = val + 1;
    return;
  }
  update(l, (l + r) / 2, L, R, n * 2, val);
  update((l + r) / 2 + 1, r, L, R, n * 2 + 1, val);
  tree[n][0] = max(tree[n * 2][0], tree[n * 2 + 1][0]);
}
long long int go(int l, int r, int L, int R, int n, long long int val) {
  if (l > R || r < L)
    return 0;
  else if (l >= L and r <= R) {
    return tree[n][0];
  }
  val = max(val, tree[n][1]);
  long long int x = max(go(l, (l + r) / 2, L, R, n * 2, val),
                        go((l + r) / 2 + 1, r, L, R, n * 2 + 1, val));
  return max(val, x);
}
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> compress, del;
  for (int i = 1; i <= m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    v[a].push_back({b, c});
    compress.push_back(b);
    compress.push_back(c);
  }
  sort(compress.begin(), compress.end());
  int sum = 0;
  for (auto it : compress) {
    if (mp[it] == 0) mp[it] = ++sum;
  }
  long long int ans = 0;
  int p = 0;
  for (int i = n; i >= 1; i--) {
    long long int mx = 0;
    for (auto it : v[i]) {
      int x = mp[it.first];
      int y = mp[it.second];
      long long int fre = go(1, sum, x, y, 1, 0);
      mx = max(mx, fre);
    }
    for (auto it : v[i]) {
      int x = mp[it.first];
      int y = mp[it.second];
      update(1, sum, x, y, 1, mx);
    }
    Max[i] = mx + 1;
    if (ans < mx + 1) {
      ans = mx + 1;
      p = i;
    }
  }
  memset(tree, 0, sizeof(tree));
  for (int i = 1; i <= n; i++) {
    if (i < p) {
      del.push_back(i);
    } else if (i == p) {
      for (auto it : v[i]) {
        update(1, sum, mp[it.first], mp[it.second], 1, 1);
      }
    } else if (i > p) {
      if (Max[i] < ans - 1) {
        del.push_back(i);
        continue;
      }
      int mx = 0;
      for (auto it : v[i]) {
        int x = mp[it.first];
        int y = mp[it.second];
        int fre = go(1, sum, x, y, 1, 0);
        mx = max(mx, fre);
      }
      if (mx) {
        for (auto it : v[p]) {
          update(1, sum, mp[it.first], mp[it.second], 1, -1);
        }
        ans--;
        for (auto it : v[i]) {
          update(1, sum, mp[it.first], mp[it.second], 1, 1);
        }
        p = i;
      } else
        del.push_back(i);
    }
  }
  cout << del.size() << "\n";
  for (auto it : del) {
    cout << it << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  int t = 1;
  while (t--) solve();
}

```