#include <bits/stdc++.h>
using namespace std;
int n, m, tot;
pair<int, int> data[300010 << 3], tag[300010 << 3];
int a[600010], nn, pre[300010], ans, ansp, used[300010];
vector<pair<int, int> > G[300010];
pair<int, int> piimax(pair<int, int> x, pair<int, int> y) {
  return x.first > y.first ? x : y;
}
void push_down(int cur) {
  if (tag[cur].first == 0) return;
  data[cur << 1] = data[cur << 1 | 1] = tag[cur];
  tag[cur << 1] = tag[cur << 1 | 1] = tag[cur];
  tag[cur] = make_pair(0, 0);
}
void push_up(int cur) {
  data[cur] = piimax(data[cur << 1], data[cur << 1 | 1]);
}
void update(int cur, int l, int r, int L, int R, pair<int, int> v) {
  if (L <= l && r <= R) {
    data[cur] = tag[cur] = v;
    return;
  }
  push_down(cur);
  int mid = (l + r) >> 1;
  if (L <= mid) update(cur << 1, l, mid, L, R, v);
  if (mid < R) update(cur << 1 | 1, mid + 1, r, L, R, v);
  push_up(cur);
}
pair<int, int> query(int cur, int l, int r, int L, int R) {
  if (L <= l && r <= R) return data[cur];
  push_down(cur);
  int mid = (l + r) >> 1;
  if (L <= mid && mid < R)
    return piimax(query(cur << 1, l, mid, L, R),
                  query(cur << 1 | 1, mid + 1, r, L, R));
  else if (L <= mid)
    return query(cur << 1, l, mid, L, R);
  else
    return query(cur << 1 | 1, mid + 1, r, L, R);
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    int id, l, r;
    scanf("%d %d %d", &id, &l, &r);
    G[id].push_back(make_pair(l, r));
    a[tot++] = l;
    a[tot++] = r;
  }
  sort(a, a + tot);
  nn = unique(a, a + tot) - a;
  for (int i = 1; i <= n; ++i) {
    int gi = G[i].size();
    for (int j = 0; j < gi; ++j) {
      G[i][j].first = lower_bound(a, a + nn, G[i][j].first) - a + 1;
      G[i][j].second = lower_bound(a, a + nn, G[i][j].second) - a + 1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    int gi = G[i].size();
    int mx = -1;
    for (int j = 0; j < gi; ++j) {
      pair<int, int> t = query(1, 1, nn, G[i][j].first, G[i][j].second);
      if (t.first > mx) {
        mx = t.first;
        pre[i] = t.second;
      }
    }
    if (mx + 1 > ans) {
      ans = mx + 1;
      ansp = i;
    }
    for (int j = 0; j < gi; ++j) {
      pair<int, int> tt = make_pair(mx + 1, i);
      update(1, 1, nn, G[i][j].first, G[i][j].second, tt);
    }
  }
  printf("%d\n", n - ans);
  for (int i = ansp; i; i = pre[i]) used[i] = 1;
  for (int i = 1; i <= n; ++i)
    if (!used[i]) printf("%d ", i);
  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;
int n, m, tot;
pair<int, int> data[300010 << 3], tag[300010 << 3];
int a[600010], nn, pre[300010], ans, ansp, used[300010];
vector<pair<int, int> > G[300010];
pair<int, int> piimax(pair<int, int> x, pair<int, int> y) {
  return x.first > y.first ? x : y;
}
void push_down(int cur) {
  if (tag[cur].first == 0) return;
  data[cur << 1] = data[cur << 1 | 1] = tag[cur];
  tag[cur << 1] = tag[cur << 1 | 1] = tag[cur];
  tag[cur] = make_pair(0, 0);
}
void push_up(int cur) {
  data[cur] = piimax(data[cur << 1], data[cur << 1 | 1]);
}
void update(int cur, int l, int r, int L, int R, pair<int, int> v) {
  if (L <= l && r <= R) {
    data[cur] = tag[cur] = v;
    return;
  }
  push_down(cur);
  int mid = (l + r) >> 1;
  if (L <= mid) update(cur << 1, l, mid, L, R, v);
  if (mid < R) update(cur << 1 | 1, mid + 1, r, L, R, v);
  push_up(cur);
}
pair<int, int> query(int cur, int l, int r, int L, int R) {
  if (L <= l && r <= R) return data[cur];
  push_down(cur);
  int mid = (l + r) >> 1;
  if (L <= mid && mid < R)
    return piimax(query(cur << 1, l, mid, L, R),
                  query(cur << 1 | 1, mid + 1, r, L, R));
  else if (L <= mid)
    return query(cur << 1, l, mid, L, R);
  else
    return query(cur << 1 | 1, mid + 1, r, L, R);
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    int id, l, r;
    scanf("%d %d %d", &id, &l, &r);
    G[id].push_back(make_pair(l, r));
    a[tot++] = l;
    a[tot++] = r;
  }
  sort(a, a + tot);
  nn = unique(a, a + tot) - a;
  for (int i = 1; i <= n; ++i) {
    int gi = G[i].size();
    for (int j = 0; j < gi; ++j) {
      G[i][j].first = lower_bound(a, a + nn, G[i][j].first) - a + 1;
      G[i][j].second = lower_bound(a, a + nn, G[i][j].second) - a + 1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    int gi = G[i].size();
    int mx = -1;
    for (int j = 0; j < gi; ++j) {
      pair<int, int> t = query(1, 1, nn, G[i][j].first, G[i][j].second);
      if (t.first > mx) {
        mx = t.first;
        pre[i] = t.second;
      }
    }
    if (mx + 1 > ans) {
      ans = mx + 1;
      ansp = i;
    }
    for (int j = 0; j < gi; ++j) {
      pair<int, int> tt = make_pair(mx + 1, i);
      update(1, 1, nn, G[i][j].first, G[i][j].second, tt);
    }
  }
  printf("%d\n", n - ans);
  for (int i = ansp; i; i = pre[i]) used[i] = 1;
  for (int i = 1; i <= n; ++i)
    if (!used[i]) printf("%d ", i);
  return 0;
}

```