#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
int n, m, opt, l, r, v, ans;
int minn[N * 40], tag[N * 40], lson[N * 40], rson[N * 40], cnt = 1;
inline void pushdown(int k) {
  if (~tag[k]) {
    minn[lson[k]] = minn[rson[k]] = tag[k];
    tag[lson[k]] = tag[rson[k]] = tag[k];
    tag[k] = -1;
  }
}
void change(int k, int l, int r, int qx, int qy, int v) {
  if (qx <= l && r <= qy) {
    minn[k] = v;
    tag[k] = v;
    return;
  }
  if (!lson[k]) lson[k] = ++cnt;
  if (!rson[k]) rson[k] = ++cnt;
  pushdown(k);
  int mid = l + r >> 1;
  if (qx <= mid) change(lson[k], l, mid, qx, qy, v);
  if (mid < qy) change(rson[k], mid + 1, r, qx, qy, v);
  minn[k] = max(minn[lson[k]], minn[rson[k]]);
}
int query(int k, int l, int r, int qx, int qy) {
  if (qx <= l && r <= qy) return minn[k];
  if (!lson[k]) lson[k] = ++cnt;
  if (!rson[k]) rson[k] = ++cnt;
  pushdown(k);
  int mid = l + r >> 1;
  int ans = 0;
  if (qx <= mid) ans = max(ans, query(lson[k], l, mid, qx, qy));
  if (mid < qy) ans = max(ans, query(rson[k], mid + 1, r, qx, qy));
  return ans;
}
vector<pair<int, int>> q[300500];
map<int, int> mp;
int nxt[300500], i, j, f[300500], mx, res[300500];
void work(int x) {
  res[x] = 1;
  int i, j, k;
  change(1, 1, 1000000000, 1, 1000000000, 0);
  for (auto [i, j] : q[x]) {
    change(1, 1, 1000000000, i, j, 1);
  }
  for (j = x + 1; j <= n; j++) {
    if (f[j] != f[x] - 1) {
      continue;
    }
    k = 0;
    for (auto [l1, r1] : q[j]) {
      k = max(k, query(1, 1, 1000000000, l1, r1));
    }
    if (k) {
      work(j);
      return;
    }
  }
}
int main() {
  memset(tag, -1, sizeof(tag));
  scanf("%d%d", &n, &m);
  for (i = 1; i <= m; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    q[x].push_back({y, z});
  }
  for (i = n; i >= 1; i--) {
    int sb = 0;
    for (auto [x, y] : q[i]) {
      sb = max(sb, query(1, 1, 1000000000, x, y));
    }
    nxt[i] = mp[sb];
    sb++;
    for (auto [x, y] : q[i]) {
      change(1, 1, 1000000000, x, y, sb);
    }
    mp[sb] = i;
    f[i] = sb;
    mx = max(mx, sb);
  }
  for (i = 1; i <= n; i++) {
    if (f[i] == mx) {
      work(i);
      break;
    }
  }
  printf("%d\n", n - mx);
  for (i = 1; i <= n; i++) {
    if (!res[i]) {
      printf("%d ", i);
    }
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 int N = 3e5 + 5;
int n, m, opt, l, r, v, ans;
int minn[N * 40], tag[N * 40], lson[N * 40], rson[N * 40], cnt = 1;
inline void pushdown(int k) {
  if (~tag[k]) {
    minn[lson[k]] = minn[rson[k]] = tag[k];
    tag[lson[k]] = tag[rson[k]] = tag[k];
    tag[k] = -1;
  }
}
void change(int k, int l, int r, int qx, int qy, int v) {
  if (qx <= l && r <= qy) {
    minn[k] = v;
    tag[k] = v;
    return;
  }
  if (!lson[k]) lson[k] = ++cnt;
  if (!rson[k]) rson[k] = ++cnt;
  pushdown(k);
  int mid = l + r >> 1;
  if (qx <= mid) change(lson[k], l, mid, qx, qy, v);
  if (mid < qy) change(rson[k], mid + 1, r, qx, qy, v);
  minn[k] = max(minn[lson[k]], minn[rson[k]]);
}
int query(int k, int l, int r, int qx, int qy) {
  if (qx <= l && r <= qy) return minn[k];
  if (!lson[k]) lson[k] = ++cnt;
  if (!rson[k]) rson[k] = ++cnt;
  pushdown(k);
  int mid = l + r >> 1;
  int ans = 0;
  if (qx <= mid) ans = max(ans, query(lson[k], l, mid, qx, qy));
  if (mid < qy) ans = max(ans, query(rson[k], mid + 1, r, qx, qy));
  return ans;
}
vector<pair<int, int>> q[300500];
map<int, int> mp;
int nxt[300500], i, j, f[300500], mx, res[300500];
void work(int x) {
  res[x] = 1;
  int i, j, k;
  change(1, 1, 1000000000, 1, 1000000000, 0);
  for (auto [i, j] : q[x]) {
    change(1, 1, 1000000000, i, j, 1);
  }
  for (j = x + 1; j <= n; j++) {
    if (f[j] != f[x] - 1) {
      continue;
    }
    k = 0;
    for (auto [l1, r1] : q[j]) {
      k = max(k, query(1, 1, 1000000000, l1, r1));
    }
    if (k) {
      work(j);
      return;
    }
  }
}
int main() {
  memset(tag, -1, sizeof(tag));
  scanf("%d%d", &n, &m);
  for (i = 1; i <= m; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    q[x].push_back({y, z});
  }
  for (i = n; i >= 1; i--) {
    int sb = 0;
    for (auto [x, y] : q[i]) {
      sb = max(sb, query(1, 1, 1000000000, x, y));
    }
    nxt[i] = mp[sb];
    sb++;
    for (auto [x, y] : q[i]) {
      change(1, 1, 1000000000, x, y, sb);
    }
    mp[sb] = i;
    f[i] = sb;
    mx = max(mx, sb);
  }
  for (i = 1; i <= n; i++) {
    if (f[i] == mx) {
      work(i);
      break;
    }
  }
  printf("%d\n", n - mx);
  for (i = 1; i <= n; i++) {
    if (!res[i]) {
      printf("%d ", i);
    }
  }
}

```