#include <bits/stdc++.h>
struct SegTree {
  struct Node {
    int l = 0, r = 0;
    std::pair<int, int> v = {0, -1}, p = {0, -1};
  };
  std::vector<Node> tree = {{}, {}};
  void visit(int n, int l, int m, int r) {
    if (!tree[n].l && l <= m) tree[n].l = tree.size(), tree.emplace_back();
    if (!tree[n].r && m + 1 <= r) tree[n].r = tree.size(), tree.emplace_back();
    setMax(tree[n].l, l, m, l, r, tree[n].p);
    setMax(tree[n].r, m + 1, r, l, r, tree[n].p);
    tree[n].p = Node().p;
  }
  void setMax(int n, int l, int r, int lq, int rq, std::pair<int, int> v) {
    if (!n || r < lq || rq < l || rq < lq) {
      return;
    } else if (lq <= l && r <= rq) {
      tree[n].v = std::max(tree[n].v, v);
      tree[n].p = std::max(tree[n].p, v);
    } else {
      int m = l + (r - l) / 2;
      visit(n, l, m, r);
      setMax(tree[n].l, l, m, lq, rq, v);
      setMax(tree[n].r, m + 1, r, lq, rq, v);
      tree[n].v = std::max(tree[tree[n].l].v, tree[tree[n].r].v);
    }
  }
  std::pair<int, int> getMax(int n, int l, int r, int lq, int rq) {
    if (!n || r < lq || rq < l || rq < lq)
      return Node().v;
    else if (lq <= l && r <= rq)
      return tree[n].v;
    int m = l + (r - l) / 2;
    visit(n, l, m, r);
    return std::max(getMax(tree[n].l, l, m, lq, rq),
                    getMax(tree[n].r, m + 1, r, lq, rq));
  }
};
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  std::map<int, int> relpoint;
  std::vector<std::vector<std::pair<int, int>>> intervals(n + 1);
  for (int x = 0, i, l, r; x < m; ++x) {
    scanf("%d %d %d", &i, &l, &r);
    intervals[i].push_back({l, r});
    relpoint[l] = relpoint[r] = 0;
  }
  int points = 0;
  for (auto &kv : relpoint) kv.second = ++points;
  SegTree tree;
  std::vector<int> bptr(n + 1, -1);
  for (int row = 1; row <= n; ++row) {
    auto &v = intervals[row];
    std::sort(v.begin(), v.end());
    std::vector<std::pair<int, int>> ranges;
    for (int i = 0, j; i < v.size(); i = j) {
      int l = v[i].first, r = v[i].second;
      for (j = i + 1; j < v.size() && v[j].first <= r; ++j)
        r = std::max(r, v[j].second);
      ranges.push_back({relpoint[l], relpoint[r]});
    }
    std::pair<int, int> dp = {0, -1};
    for (auto &lr : ranges)
      dp = std::max(dp, tree.getMax(1, 1, points, lr.first, lr.second));
    for (auto &lr : ranges) {
      tree.setMax(1, 1, points, lr.first, lr.second, {dp.first + 1, row});
    }
    bptr[row] = dp.second;
  }
  auto ans = tree.getMax(1, 1, points, 1, points);
  printf("%d\n", n - ans.first);
  for (int i = ans.second, j; i != -1;) i = bptr[j = i], bptr[j] = 0;
  for (int row = 1; row <= n; ++row)
    if (bptr[row] != 0) printf("%d ", row);
  if (ans.first != n) printf("\n");
}

### 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>
struct SegTree {
  struct Node {
    int l = 0, r = 0;
    std::pair<int, int> v = {0, -1}, p = {0, -1};
  };
  std::vector<Node> tree = {{}, {}};
  void visit(int n, int l, int m, int r) {
    if (!tree[n].l && l <= m) tree[n].l = tree.size(), tree.emplace_back();
    if (!tree[n].r && m + 1 <= r) tree[n].r = tree.size(), tree.emplace_back();
    setMax(tree[n].l, l, m, l, r, tree[n].p);
    setMax(tree[n].r, m + 1, r, l, r, tree[n].p);
    tree[n].p = Node().p;
  }
  void setMax(int n, int l, int r, int lq, int rq, std::pair<int, int> v) {
    if (!n || r < lq || rq < l || rq < lq) {
      return;
    } else if (lq <= l && r <= rq) {
      tree[n].v = std::max(tree[n].v, v);
      tree[n].p = std::max(tree[n].p, v);
    } else {
      int m = l + (r - l) / 2;
      visit(n, l, m, r);
      setMax(tree[n].l, l, m, lq, rq, v);
      setMax(tree[n].r, m + 1, r, lq, rq, v);
      tree[n].v = std::max(tree[tree[n].l].v, tree[tree[n].r].v);
    }
  }
  std::pair<int, int> getMax(int n, int l, int r, int lq, int rq) {
    if (!n || r < lq || rq < l || rq < lq)
      return Node().v;
    else if (lq <= l && r <= rq)
      return tree[n].v;
    int m = l + (r - l) / 2;
    visit(n, l, m, r);
    return std::max(getMax(tree[n].l, l, m, lq, rq),
                    getMax(tree[n].r, m + 1, r, lq, rq));
  }
};
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  std::map<int, int> relpoint;
  std::vector<std::vector<std::pair<int, int>>> intervals(n + 1);
  for (int x = 0, i, l, r; x < m; ++x) {
    scanf("%d %d %d", &i, &l, &r);
    intervals[i].push_back({l, r});
    relpoint[l] = relpoint[r] = 0;
  }
  int points = 0;
  for (auto &kv : relpoint) kv.second = ++points;
  SegTree tree;
  std::vector<int> bptr(n + 1, -1);
  for (int row = 1; row <= n; ++row) {
    auto &v = intervals[row];
    std::sort(v.begin(), v.end());
    std::vector<std::pair<int, int>> ranges;
    for (int i = 0, j; i < v.size(); i = j) {
      int l = v[i].first, r = v[i].second;
      for (j = i + 1; j < v.size() && v[j].first <= r; ++j)
        r = std::max(r, v[j].second);
      ranges.push_back({relpoint[l], relpoint[r]});
    }
    std::pair<int, int> dp = {0, -1};
    for (auto &lr : ranges)
      dp = std::max(dp, tree.getMax(1, 1, points, lr.first, lr.second));
    for (auto &lr : ranges) {
      tree.setMax(1, 1, points, lr.first, lr.second, {dp.first + 1, row});
    }
    bptr[row] = dp.second;
  }
  auto ans = tree.getMax(1, 1, points, 1, points);
  printf("%d\n", n - ans.first);
  for (int i = ans.second, j; i != -1;) i = bptr[j = i], bptr[j] = 0;
  for (int row = 1; row <= n; ++row)
    if (bptr[row] != 0) printf("%d ", row);
  if (ans.first != n) printf("\n");
}

```