#include <bits/stdc++.h>
using namespace std;
struct segment {
  int l, u;
} tmp;
pair<int, int> seg[4 * int(6e5 + 10)], lazy[4 * int(6e5 + 10)], x;
int p[int(6e5 + 10)], n, m, k, dp;
pair<int, int> ans = {-1, -1};
vector<segment> g[int(6e5 + 10)];
set<int> s;
map<int, int> mp;
bool vis[int(6e5 + 10)];
pair<int, int> fmax(pair<int, int> a, pair<int, int> b) {
  if (a.first < b.first) return b;
  return a;
}
void build(int l, int h, int pos) {
  if (l == h) {
    seg[pos] = lazy[pos] = {0, 0};
    return;
  }
  int mid = (l + h) / 2;
  build(l, mid, 2 * pos);
  build(mid + 1, h, 2 * pos + 1);
  seg[pos] = lazy[pos] = {0, 0};
}
void update(int l, int h, int x, int y, pair<int, int> z, int pos) {
  if (lazy[pos].first) {
    seg[pos] = fmax(seg[pos], lazy[pos]);
    if (l != h) {
      lazy[2 * pos] = fmax(lazy[pos], lazy[2 * pos]);
      lazy[2 * pos + 1] = fmax(lazy[pos], lazy[2 * pos + 1]);
    }
    lazy[pos] = {0, 0};
  }
  if (h < x || l > y) return;
  if (x <= l && h <= y) {
    seg[pos] = fmax(seg[pos], z);
    if (l != h) {
      lazy[2 * pos] = fmax(lazy[2 * pos], z);
      lazy[2 * pos + 1] = fmax(lazy[2 * pos], z);
    }
    return;
  }
  int mid = (l + h) / 2;
  update(l, mid, x, y, z, 2 * pos);
  update(mid + 1, h, x, y, z, 2 * pos + 1);
  seg[pos] = fmax(seg[2 * pos], seg[2 * pos + 1]);
}
pair<int, int> query(int l, int h, int x, int y, int pos) {
  if (lazy[pos].first) {
    seg[pos] = fmax(seg[pos], lazy[pos]);
    if (l != h) {
      lazy[2 * pos] = fmax(lazy[pos], lazy[2 * pos]);
      lazy[2 * pos + 1] = fmax(lazy[pos], lazy[2 * pos + 1]);
    }
    lazy[pos] = {0, 0};
  }
  if (h < x || l > y) return {-1, -1};
  if (x <= l && h <= y) {
    return seg[pos];
  }
  int mid = (l + h) / 2;
  return fmax(query(l, mid, x, y, 2 * pos),
              query(mid + 1, h, x, y, 2 * pos + 1));
}
void fprint(int t) {
  if (p[t] != t) {
    fprint(p[t]);
  }
  vis[t] = true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout.precision(10);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> k >> tmp.l >> tmp.u;
    s.insert(tmp.l);
    s.insert(tmp.u);
    g[k].push_back(tmp);
  }
  k = 1;
  for (auto x : s) {
    if (mp.find(x) == mp.end()) {
      mp[x] = k++;
    }
  }
  build(1, k, 1);
  for (int i = 1; i <= n; i++) {
    p[i] = i;
    dp = 1;
    for (auto s : g[i]) {
      x = query(1, k, mp[s.l], mp[s.u], 1);
      if (dp < 1 + x.first) {
        p[i] = x.second;
        dp = 1 + x.first;
      }
    }
    for (auto s : g[i]) {
      update(1, k, mp[s.l], mp[s.u], make_pair(dp, i), 1);
    }
    ans = fmax(ans, make_pair(dp, i));
  }
  cout << n - ans.first << endl;
  fprint(ans.second);
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) cout << i << ' ';
  }
  cout << endl;
  return 0;
}

### Prompt

Please formulate 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;
struct segment {
  int l, u;
} tmp;
pair<int, int> seg[4 * int(6e5 + 10)], lazy[4 * int(6e5 + 10)], x;
int p[int(6e5 + 10)], n, m, k, dp;
pair<int, int> ans = {-1, -1};
vector<segment> g[int(6e5 + 10)];
set<int> s;
map<int, int> mp;
bool vis[int(6e5 + 10)];
pair<int, int> fmax(pair<int, int> a, pair<int, int> b) {
  if (a.first < b.first) return b;
  return a;
}
void build(int l, int h, int pos) {
  if (l == h) {
    seg[pos] = lazy[pos] = {0, 0};
    return;
  }
  int mid = (l + h) / 2;
  build(l, mid, 2 * pos);
  build(mid + 1, h, 2 * pos + 1);
  seg[pos] = lazy[pos] = {0, 0};
}
void update(int l, int h, int x, int y, pair<int, int> z, int pos) {
  if (lazy[pos].first) {
    seg[pos] = fmax(seg[pos], lazy[pos]);
    if (l != h) {
      lazy[2 * pos] = fmax(lazy[pos], lazy[2 * pos]);
      lazy[2 * pos + 1] = fmax(lazy[pos], lazy[2 * pos + 1]);
    }
    lazy[pos] = {0, 0};
  }
  if (h < x || l > y) return;
  if (x <= l && h <= y) {
    seg[pos] = fmax(seg[pos], z);
    if (l != h) {
      lazy[2 * pos] = fmax(lazy[2 * pos], z);
      lazy[2 * pos + 1] = fmax(lazy[2 * pos], z);
    }
    return;
  }
  int mid = (l + h) / 2;
  update(l, mid, x, y, z, 2 * pos);
  update(mid + 1, h, x, y, z, 2 * pos + 1);
  seg[pos] = fmax(seg[2 * pos], seg[2 * pos + 1]);
}
pair<int, int> query(int l, int h, int x, int y, int pos) {
  if (lazy[pos].first) {
    seg[pos] = fmax(seg[pos], lazy[pos]);
    if (l != h) {
      lazy[2 * pos] = fmax(lazy[pos], lazy[2 * pos]);
      lazy[2 * pos + 1] = fmax(lazy[pos], lazy[2 * pos + 1]);
    }
    lazy[pos] = {0, 0};
  }
  if (h < x || l > y) return {-1, -1};
  if (x <= l && h <= y) {
    return seg[pos];
  }
  int mid = (l + h) / 2;
  return fmax(query(l, mid, x, y, 2 * pos),
              query(mid + 1, h, x, y, 2 * pos + 1));
}
void fprint(int t) {
  if (p[t] != t) {
    fprint(p[t]);
  }
  vis[t] = true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout.precision(10);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> k >> tmp.l >> tmp.u;
    s.insert(tmp.l);
    s.insert(tmp.u);
    g[k].push_back(tmp);
  }
  k = 1;
  for (auto x : s) {
    if (mp.find(x) == mp.end()) {
      mp[x] = k++;
    }
  }
  build(1, k, 1);
  for (int i = 1; i <= n; i++) {
    p[i] = i;
    dp = 1;
    for (auto s : g[i]) {
      x = query(1, k, mp[s.l], mp[s.u], 1);
      if (dp < 1 + x.first) {
        p[i] = x.second;
        dp = 1 + x.first;
      }
    }
    for (auto s : g[i]) {
      update(1, k, mp[s.l], mp[s.u], make_pair(dp, i), 1);
    }
    ans = fmax(ans, make_pair(dp, i));
  }
  cout << n - ans.first << endl;
  fprint(ans.second);
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) cout << i << ' ';
  }
  cout << endl;
  return 0;
}

```