#include <bits/stdc++.h>
using namespace std;
struct gg {
  int c, fa;
  bool operator<(const gg &QAQ) const { return c < QAQ.c; }
} g0, ans;
struct lxy {
  int lson, rson;
  gg v, tg;
} b[2000005];
struct lxy1 {
  int x, l, r;
  bool operator<(const lxy1 &QAQ) const { return x < QAQ.x; }
} q[300005];
int n, m, cnt, rt, N;
int g[300005];
bool vis[300005];
priority_queue<int> p;
map<int, int> h;
void update(int u) { b[u].v = max(b[b[u].lson].v, b[b[u].rson].v); }
void pushdown(int u) {
  if (b[u].tg.c == 0) return;
  b[b[u].lson].v = b[u].tg, b[b[u].lson].tg = b[u].tg;
  b[b[u].rson].v = b[u].tg, b[b[u].rson].tg = b[u].tg;
  b[u].tg.c = 0;
}
void build(int &u, int l, int r) {
  u = ++cnt;
  if (l == r) return;
  int mid = (l + r) / 2;
  build(b[u].lson, l, mid);
  build(b[u].rson, mid + 1, r);
}
void modify(int u, int L, int R, int l, int r, int k, int id) {
  if (L == l && R == r) {
    b[u].v.c = k, b[u].tg.c = k;
    b[u].v.fa = id, b[u].tg.fa = id;
    return;
  }
  int mid = (L + R) / 2;
  if (mid >= r)
    pushdown(u), modify(b[u].lson, L, mid, l, r, k, id);
  else if (l > mid)
    pushdown(u), modify(b[u].rson, mid + 1, R, l, r, k, id);
  else
    pushdown(u), modify(b[u].lson, L, mid, l, mid, k, id),
        modify(b[u].rson, mid + 1, R, mid + 1, r, k, id);
  update(u);
}
gg query(int u, int L, int R, int l, int r) {
  pushdown(u);
  if (L == l && R == r) return b[u].v;
  int mid = (L + R) / 2;
  if (mid >= r) {
    return query(b[u].lson, L, mid, l, r);
  } else if (l > mid) {
    return query(b[u].rson, mid + 1, R, l, r);
  } else {
    return max(query(b[u].lson, L, mid, l, mid),
               query(b[u].rson, mid + 1, R, mid + 1, r));
  }
  update(u);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &q[i].x, &q[i].l, &q[i].r);
    p.push(-q[i].l), p.push(-q[i].r);
  }
  sort(q + 1, q + 1 + m);
  N = p.size();
  build(rt, 1, p.size());
  cnt = 0;
  while (!p.empty()) {
    int y = p.top();
    h[-p.top()] = ++cnt;
    p.pop();
    while (!p.empty() && p.top() == y) p.pop();
  }
  for (int i = 1; i <= m; i++) {
    gg ret = g0;
    int now = i;
    while (q[now].x == q[now + 1].x) {
      ret = max(ret, query(rt, 1, N, h[q[now].l], h[q[now].r]));
      now++;
    }
    ret = max(ret, query(rt, 1, N, h[q[now].l], h[q[now].r]));
    for (int j = i; j <= now; j++)
      modify(rt, 1, N, h[q[j].l], h[q[j].r], ret.c + 1, q[j].x);
    i = now;
    g[q[i].x] = ret.fa;
    if (ret.c + 1 > ans.c) ans.c = ret.c + 1, ans.fa = q[i].x;
  }
  int pos = ans.fa;
  while (pos != 0) {
    vis[pos] = 1;
    pos = g[pos];
  }
  cout << n - ans.c << endl;
  for (int i = 1; i <= n; i++)
    if (vis[i] == 0) printf("%d ", i);
}

### Prompt

In cpp, your task is to solve 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 gg {
  int c, fa;
  bool operator<(const gg &QAQ) const { return c < QAQ.c; }
} g0, ans;
struct lxy {
  int lson, rson;
  gg v, tg;
} b[2000005];
struct lxy1 {
  int x, l, r;
  bool operator<(const lxy1 &QAQ) const { return x < QAQ.x; }
} q[300005];
int n, m, cnt, rt, N;
int g[300005];
bool vis[300005];
priority_queue<int> p;
map<int, int> h;
void update(int u) { b[u].v = max(b[b[u].lson].v, b[b[u].rson].v); }
void pushdown(int u) {
  if (b[u].tg.c == 0) return;
  b[b[u].lson].v = b[u].tg, b[b[u].lson].tg = b[u].tg;
  b[b[u].rson].v = b[u].tg, b[b[u].rson].tg = b[u].tg;
  b[u].tg.c = 0;
}
void build(int &u, int l, int r) {
  u = ++cnt;
  if (l == r) return;
  int mid = (l + r) / 2;
  build(b[u].lson, l, mid);
  build(b[u].rson, mid + 1, r);
}
void modify(int u, int L, int R, int l, int r, int k, int id) {
  if (L == l && R == r) {
    b[u].v.c = k, b[u].tg.c = k;
    b[u].v.fa = id, b[u].tg.fa = id;
    return;
  }
  int mid = (L + R) / 2;
  if (mid >= r)
    pushdown(u), modify(b[u].lson, L, mid, l, r, k, id);
  else if (l > mid)
    pushdown(u), modify(b[u].rson, mid + 1, R, l, r, k, id);
  else
    pushdown(u), modify(b[u].lson, L, mid, l, mid, k, id),
        modify(b[u].rson, mid + 1, R, mid + 1, r, k, id);
  update(u);
}
gg query(int u, int L, int R, int l, int r) {
  pushdown(u);
  if (L == l && R == r) return b[u].v;
  int mid = (L + R) / 2;
  if (mid >= r) {
    return query(b[u].lson, L, mid, l, r);
  } else if (l > mid) {
    return query(b[u].rson, mid + 1, R, l, r);
  } else {
    return max(query(b[u].lson, L, mid, l, mid),
               query(b[u].rson, mid + 1, R, mid + 1, r));
  }
  update(u);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &q[i].x, &q[i].l, &q[i].r);
    p.push(-q[i].l), p.push(-q[i].r);
  }
  sort(q + 1, q + 1 + m);
  N = p.size();
  build(rt, 1, p.size());
  cnt = 0;
  while (!p.empty()) {
    int y = p.top();
    h[-p.top()] = ++cnt;
    p.pop();
    while (!p.empty() && p.top() == y) p.pop();
  }
  for (int i = 1; i <= m; i++) {
    gg ret = g0;
    int now = i;
    while (q[now].x == q[now + 1].x) {
      ret = max(ret, query(rt, 1, N, h[q[now].l], h[q[now].r]));
      now++;
    }
    ret = max(ret, query(rt, 1, N, h[q[now].l], h[q[now].r]));
    for (int j = i; j <= now; j++)
      modify(rt, 1, N, h[q[j].l], h[q[j].r], ret.c + 1, q[j].x);
    i = now;
    g[q[i].x] = ret.fa;
    if (ret.c + 1 > ans.c) ans.c = ret.c + 1, ans.fa = q[i].x;
  }
  int pos = ans.fa;
  while (pos != 0) {
    vis[pos] = 1;
    pos = g[pos];
  }
  cout << n - ans.c << endl;
  for (int i = 1; i <= n; i++)
    if (vis[i] == 0) printf("%d ", i);
}

```