#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read() {
  int ret = 0, sgn = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') sgn = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * sgn;
}
inline void Out(int a) {
  if (a > 9) Out(a / 10);
  putchar(a % 10 + '0');
}
long long __gcd(long long a, long long b) {
  return b == 0 ? a : __gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a * b / __gcd(a, b); }
long long qpow(int m, int k, int mod) {
  long long res = 1, t = m;
  while (k) {
    if (k & 1) res = res * t % mod;
    t = t * t % mod;
    k >>= 1;
  }
  return res;
}
int Fermat(int a, int p) { return qpow(a, p - 2, p); }
int exgcd(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int g = exgcd(b, a % b, x, y);
  int t = x;
  x = y;
  y = t - a / b * y;
  return g;
}
int mod_reverse(int a, int p) {
  int d, x, y;
  d = exgcd(a, p, x, y);
  if (d == 1)
    return (x % p + p) % p;
  else
    return -1;
}
long long china(int a[], int b[], int n) {
  int M = 1, y, x = 0;
  for (int i = 0; i < n; ++i) M *= a[i];
  for (int i = 0; i < n; ++i) {
    int w = M / a[i];
    int tx = 0;
    int t = exgcd(w, a[i], tx, y);
    x = (x + w * (b[i] / t) * x) % M;
  }
  return (x + M) % M;
}
struct node {
  int l, r;
  int id;
} a[200005];
int n, k;
int pos, maxn, minn;
int cmp(node a, node b) { return a.l < b.l; }
vector<int> ans;
set<pair<int, int> > st;
int main() {
  scanf("%d%d", &n, &k);
  maxn, minn;
  pair<int, int> tmp;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &a[i].l, &a[i].r);
    a[i].id = i;
    maxn = max(maxn, a[i].r);
    minn = min(minn, a[i].l);
  }
  sort(a + 1, a + 1 + n, cmp);
  pos = 1;
  for (int i = minn; i <= maxn; i++) {
    while (pos <= n && a[pos].l <= i) {
      st.insert(make_pair(a[pos].r, a[pos].id));
      pos++;
    }
    while (st.size() && st.begin()->first < i) {
      st.erase(st.begin());
    }
    while (st.size() > k) {
      tmp = *(--st.end());
      ans.push_back(tmp.second);
      st.erase(tmp);
    }
  }
  sort(ans.begin(), ans.end());
  printf("%d\n", ans.size());
  for (auto v : ans) {
    printf("%d ", v);
  }
  printf("\n");
}

### Prompt

Your task is to create a cpp solution to the following problem:

The only difference between easy and hard versions is constraints.

You are given n segments on the coordinate axis OX. Segments can intersect, lie inside each other and even coincide. The i-th segment is [l_i; r_i] (l_i ≤ r_i) and it covers all integer points j such that l_i ≤ j ≤ r_i.

The integer point is called bad if it is covered by strictly more than k segments.

Your task is to remove the minimum number of segments so that there are no bad points at all.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the number of segments and the maximum number of segments by which each integer point can be covered.

The next n lines contain segments. The i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 2 ⋅ 10^5) — the endpoints of the i-th segment.

Output

In the first line print one integer m (0 ≤ m ≤ n) — the minimum number of segments you need to remove so that there are no bad points.

In the second line print m distinct integers p_1, p_2, ..., p_m (1 ≤ p_i ≤ n) — indices of segments you remove in any order. If there are multiple answers, you can print any of them.

Examples

Input


7 2
11 11
9 11
7 8
8 9
7 8
9 11
7 9


Output


3
4 6 7 


Input


5 1
29 30
30 30
29 29
28 30
30 30


Output


3
1 4 5 


Input


6 1
2 3
3 3
2 3
2 2
2 3
2 3


Output


4
1 3 5 6 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read() {
  int ret = 0, sgn = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') sgn = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * sgn;
}
inline void Out(int a) {
  if (a > 9) Out(a / 10);
  putchar(a % 10 + '0');
}
long long __gcd(long long a, long long b) {
  return b == 0 ? a : __gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a * b / __gcd(a, b); }
long long qpow(int m, int k, int mod) {
  long long res = 1, t = m;
  while (k) {
    if (k & 1) res = res * t % mod;
    t = t * t % mod;
    k >>= 1;
  }
  return res;
}
int Fermat(int a, int p) { return qpow(a, p - 2, p); }
int exgcd(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int g = exgcd(b, a % b, x, y);
  int t = x;
  x = y;
  y = t - a / b * y;
  return g;
}
int mod_reverse(int a, int p) {
  int d, x, y;
  d = exgcd(a, p, x, y);
  if (d == 1)
    return (x % p + p) % p;
  else
    return -1;
}
long long china(int a[], int b[], int n) {
  int M = 1, y, x = 0;
  for (int i = 0; i < n; ++i) M *= a[i];
  for (int i = 0; i < n; ++i) {
    int w = M / a[i];
    int tx = 0;
    int t = exgcd(w, a[i], tx, y);
    x = (x + w * (b[i] / t) * x) % M;
  }
  return (x + M) % M;
}
struct node {
  int l, r;
  int id;
} a[200005];
int n, k;
int pos, maxn, minn;
int cmp(node a, node b) { return a.l < b.l; }
vector<int> ans;
set<pair<int, int> > st;
int main() {
  scanf("%d%d", &n, &k);
  maxn, minn;
  pair<int, int> tmp;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &a[i].l, &a[i].r);
    a[i].id = i;
    maxn = max(maxn, a[i].r);
    minn = min(minn, a[i].l);
  }
  sort(a + 1, a + 1 + n, cmp);
  pos = 1;
  for (int i = minn; i <= maxn; i++) {
    while (pos <= n && a[pos].l <= i) {
      st.insert(make_pair(a[pos].r, a[pos].id));
      pos++;
    }
    while (st.size() && st.begin()->first < i) {
      st.erase(st.begin());
    }
    while (st.size() > k) {
      tmp = *(--st.end());
      ans.push_back(tmp.second);
      st.erase(tmp);
    }
  }
  sort(ans.begin(), ans.end());
  printf("%d\n", ans.size());
  for (auto v : ans) {
    printf("%d ", v);
  }
  printf("\n");
}

```