#include <bits/stdc++.h>
using namespace std;
template <int maxn>
struct lazy {
  pair<int, int> tree[maxn * 4 + 5], pend[maxn * 4 + 5];
  lazy() {
    for (int i = 0; i < maxn * 4 + 3; ++i)
      tree[i] = {1000000000, 1000000000}, pend[i] = {1000000000, 1000000000};
  }
  void propagate(int cur, int ss, int se) {
    if (pend[cur].first != 1000000000) {
      tree[cur] = min(tree[cur], pend[cur]);
      if (ss != se) {
        pend[cur * 2 + 1] = min(pend[cur * 2 + 1], pend[cur]);
        pend[cur * 2 + 2] = min(pend[cur * 2 + 2], pend[cur]);
      }
      pend[cur].first = 1000000000;
    }
  }
  void addUtil(int cur, int ss, int se, int l, int r, pair<int, int> val) {
    propagate(cur, ss, se);
    if (ss > se || ss > r || se < l) return;
    if (ss >= l && se <= r) {
      pend[cur] = val;
      propagate(cur, ss, se);
      return;
    }
    addUtil(cur * 2 + 1, ss, (ss + se) / 2, l, r, val);
    addUtil(cur * 2 + 2, (ss + se) / 2 + 1, se, l, r, val);
    tree[cur] = min(tree[cur * 2 + 1], tree[cur * 2 + 2]);
  }
  void add(int l, int r, pair<int, int> val) {
    addUtil(0, 0, maxn - 1, l, r, val);
  }
  pair<int, int> getUtil(int ss, int se, int l, int r, int cur) {
    propagate(cur, ss, se);
    if (ss > se || ss > r || se < l) return {1000000000, 1000000000};
    if (ss >= l && se <= r) {
      return tree[cur];
    }
    return min(getUtil(ss, (ss + se) / 2, l, r, cur * 2 + 1),
               getUtil((ss + se) / 2 + 1, se, l, r, cur * 2 + 2));
  }
  pair<int, int> get(int l, int r) { return getUtil(0, maxn - 1, l, r, 0); }
};
lazy<600005> dps;
int dp[300005];
vector<pair<int, int>> seg[300005];
int back[300005];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    int j, l, r;
    cin >> j >> l >> r;
    --j;
    seg[j].push_back({l, r});
  }
  for (int i = 0; i < n; ++i) sort(seg[i].begin(), seg[i].end());
  set<int> all;
  for (int i = 0; i < n; ++i) {
    for (auto j : seg[i]) all.insert(j.first), all.insert(j.second);
  }
  map<int, int> indx;
  int k = 0;
  for (auto i : all) {
    indx[i] = k++;
  }
  for (int i = 0; i < n; ++i) {
    for (auto& j : seg[i]) {
      j.first = indx[j.first];
      j.second = indx[j.second];
    }
  }
  dp[n - 1] = 0;
  back[n - 1] = -1;
  for (auto j : seg[n - 1]) {
    auto [l, r] = j;
    dps.add(l, r, {n - 1, n - 1});
  }
  for (int i = n - 2; i >= 0; --i) {
    int& res = dp[i];
    res = n - i - 1;
    back[i] = -1;
    for (auto j : seg[i]) {
      auto [l, r] = j;
      if (res > dps.get(l, r).first - (i + 1)) {
        back[i] = dps.get(l, r).second;
      }
      res = min(res, dps.get(l, r).first - (i + 1));
    }
    for (auto j : seg[i]) {
      auto [l, r] = j;
      dps.add(l, r, {res + i, i});
    }
  }
  int ans = 1000000000;
  for (int i = 0; i < n; ++i) ans = min(ans, dp[i] + i);
  assert(ans != 1000000000);
  cout << ans << endl;
  for (int i = 0; i < n; ++i) {
    if (ans == dp[i] + i) {
      int cur = i;
      vector<int> all;
      while (back[cur] != -1) {
        all.push_back(cur);
        cur = back[cur];
      }
      all.push_back(cur);
      for (int j = 0; j < i; ++j) cout << j + 1 << ' ';
      for (int j = 0; j + 1 < all.size(); ++j) {
        for (int k = all[j] + 1; k < all[j + 1]; ++k) {
          cout << k + 1 << ' ';
        }
      }
      if (!all.empty())
        for (int j = all.back() + 1; j < n; ++j) cout << j + 1 << ' ';
      cout << endl;
      return 0;
    }
  }
  assert(0);
}

### 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;
template <int maxn>
struct lazy {
  pair<int, int> tree[maxn * 4 + 5], pend[maxn * 4 + 5];
  lazy() {
    for (int i = 0; i < maxn * 4 + 3; ++i)
      tree[i] = {1000000000, 1000000000}, pend[i] = {1000000000, 1000000000};
  }
  void propagate(int cur, int ss, int se) {
    if (pend[cur].first != 1000000000) {
      tree[cur] = min(tree[cur], pend[cur]);
      if (ss != se) {
        pend[cur * 2 + 1] = min(pend[cur * 2 + 1], pend[cur]);
        pend[cur * 2 + 2] = min(pend[cur * 2 + 2], pend[cur]);
      }
      pend[cur].first = 1000000000;
    }
  }
  void addUtil(int cur, int ss, int se, int l, int r, pair<int, int> val) {
    propagate(cur, ss, se);
    if (ss > se || ss > r || se < l) return;
    if (ss >= l && se <= r) {
      pend[cur] = val;
      propagate(cur, ss, se);
      return;
    }
    addUtil(cur * 2 + 1, ss, (ss + se) / 2, l, r, val);
    addUtil(cur * 2 + 2, (ss + se) / 2 + 1, se, l, r, val);
    tree[cur] = min(tree[cur * 2 + 1], tree[cur * 2 + 2]);
  }
  void add(int l, int r, pair<int, int> val) {
    addUtil(0, 0, maxn - 1, l, r, val);
  }
  pair<int, int> getUtil(int ss, int se, int l, int r, int cur) {
    propagate(cur, ss, se);
    if (ss > se || ss > r || se < l) return {1000000000, 1000000000};
    if (ss >= l && se <= r) {
      return tree[cur];
    }
    return min(getUtil(ss, (ss + se) / 2, l, r, cur * 2 + 1),
               getUtil((ss + se) / 2 + 1, se, l, r, cur * 2 + 2));
  }
  pair<int, int> get(int l, int r) { return getUtil(0, maxn - 1, l, r, 0); }
};
lazy<600005> dps;
int dp[300005];
vector<pair<int, int>> seg[300005];
int back[300005];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    int j, l, r;
    cin >> j >> l >> r;
    --j;
    seg[j].push_back({l, r});
  }
  for (int i = 0; i < n; ++i) sort(seg[i].begin(), seg[i].end());
  set<int> all;
  for (int i = 0; i < n; ++i) {
    for (auto j : seg[i]) all.insert(j.first), all.insert(j.second);
  }
  map<int, int> indx;
  int k = 0;
  for (auto i : all) {
    indx[i] = k++;
  }
  for (int i = 0; i < n; ++i) {
    for (auto& j : seg[i]) {
      j.first = indx[j.first];
      j.second = indx[j.second];
    }
  }
  dp[n - 1] = 0;
  back[n - 1] = -1;
  for (auto j : seg[n - 1]) {
    auto [l, r] = j;
    dps.add(l, r, {n - 1, n - 1});
  }
  for (int i = n - 2; i >= 0; --i) {
    int& res = dp[i];
    res = n - i - 1;
    back[i] = -1;
    for (auto j : seg[i]) {
      auto [l, r] = j;
      if (res > dps.get(l, r).first - (i + 1)) {
        back[i] = dps.get(l, r).second;
      }
      res = min(res, dps.get(l, r).first - (i + 1));
    }
    for (auto j : seg[i]) {
      auto [l, r] = j;
      dps.add(l, r, {res + i, i});
    }
  }
  int ans = 1000000000;
  for (int i = 0; i < n; ++i) ans = min(ans, dp[i] + i);
  assert(ans != 1000000000);
  cout << ans << endl;
  for (int i = 0; i < n; ++i) {
    if (ans == dp[i] + i) {
      int cur = i;
      vector<int> all;
      while (back[cur] != -1) {
        all.push_back(cur);
        cur = back[cur];
      }
      all.push_back(cur);
      for (int j = 0; j < i; ++j) cout << j + 1 << ' ';
      for (int j = 0; j + 1 < all.size(); ++j) {
        for (int k = all[j] + 1; k < all[j + 1]; ++k) {
          cout << k + 1 << ' ';
        }
      }
      if (!all.empty())
        for (int j = all.back() + 1; j < n; ++j) cout << j + 1 << ' ';
      cout << endl;
      return 0;
    }
  }
  assert(0);
}

```