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 ≤ 200) — 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 ≤ 200) — 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
1 4 7 


Input


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


Output


3
1 2 4 


Input


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


Output


4
1 3 5 6 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
void chmax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
void chmin(T &a, T b) {
  if (a > b) a = b;
}
constexpr long long INF = 1000000000000000000;
constexpr long long mod = 1000000007;
constexpr long double eps = 1e-8;
const long double pi = acos(-1);
long long dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
long long Random(long long mi, long long ma) {
  random_device rnd;
  mt19937 mt(rnd());
  uniform_int_distribution<long long> engine(mi, ma);
  return engine(mt);
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
bool prime(long long a) {
  if (a == 1) return false;
  for (long long i = 2; i * i <= a; i++) {
    if (a % i == 0) return false;
  }
  return true;
}
long long modpow(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2) return modpow(a, b - 1) * a % mod;
  long long memo = modpow(a, b / 2);
  return memo * memo % mod;
}
vector<long long> kaijo, invkaijo;
void init_fact(long long n) {
  kaijo.resize(n + 1);
  invkaijo.resize(n + 1);
  kaijo[0] = 1;
  for (long long i = 1; i <= n; i++) {
    kaijo[i] = kaijo[i - 1] * i;
    kaijo[i] %= mod;
  }
  for (long long i = 0; i < n + 1; i++) invkaijo[i] = modpow(kaijo[i], mod - 2);
}
long long comb(long long a, long long b) {
  if (a < b) return 0;
  if (a < 0 || b < 0) return 0;
  return kaijo[a] * modpow(kaijo[a - b], mod - 2) % mod *
         modpow(kaijo[b], mod - 2) % mod;
}
long long inv(long long x) {
  x = modpow(x, mod - 2);
  return x;
}
struct BIT {
  long long n, beki = 1;
  vector<long long> bit;
  BIT(long long x) {
    bit.resize(x + 1, 0);
    n = x;
    while (beki * 2 <= n) beki *= 2;
  }
  long long sum(long long i) {
    long long res = 0;
    while (i > 0) {
      res += bit[i];
      i -= i & -i;
    }
    return res;
  }
  void add(long long i, long long x) {
    while (i <= n) {
      bit[i] += x;
      i += i & -i;
    }
  }
  long long lowerbound(long long w) {
    if (w <= 0) return 0;
    long long x = 0;
    for (long long k = beki; k > 0; k >>= 1) {
      if (x + k <= n && bit[x + k] < w) {
        w -= bit[x + k];
        x += k;
      }
    }
    return x + 1;
  }
};
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n, k;
  cin >> n >> k;
  vector<pair<long long, long long> > v[200010];
  BIT bit(200001);
  for (long long i = 0; i < n; i++) {
    long long l, r;
    cin >> l >> r;
    bit.add(l, 1);
    bit.add(r + 1, -1);
    v[l].push_back({r, i + 1});
  }
  priority_queue<pair<long long, long long> > q;
  vector<long long> ans;
  for (long long i = 1; i <= 200000; i++) {
    for (pair<long long, long long> j : v[i]) q.push(j);
    while (bit.sum(i) > k) {
      bit.add(i, -1);
      bit.add(q.top().first + 1, 1);
      ans.push_back(q.top().second);
      q.pop();
    }
  }
  cout << ans.size() << endl;
  for (long long i : ans) cout << i << " ";
  cout << endl;
  return 0;
}