You are given a permutation P_1 ... P_N of the set {1, 2, ..., N}.

You can apply the following operation to this permutation, any number of times (possibly zero):

* Choose two indices i,j (1 ≦ i < j ≦ N), such that j - i ≧ K and |P_i - P_j| = 1. Then, swap the values of P_i and P_j.



Among all permutations that can be obtained by applying this operation to the given permutation, find the lexicographically smallest one.

Constraints

* 2≦N≦500,000
* 1≦K≦N-1
* P is a permutation of the set {1, 2, ..., N}.

Input

The input is given from Standard Input in the following format:


N K
P_1 P_2 ... P_N


Output

Print the lexicographically smallest permutation that can be obtained.

Examples

Input

4 2
4 2 3 1


Output

2
1
4
3


Input

5 1
5 4 3 2 1


Output

1
2
3
4
5


Input

8 3
4 5 7 8 3 1 2 6


Output

1
2
6
7
5
3
4
8
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long n, a[300010], b[300010];
long long val[4100][4100], ans;
long long fac[9010], inv[9010];
void add(long long &x, long long y) {
  x += y;
  if (x >= 1000000007) x -= 1000000007;
  while (x < 0) x += 1000000007;
}
inline long long qpow(long long bas, long long pw) {
  long long ans = 1;
  for (; pw; pw >>= 1, (bas *= bas) %= 1000000007)
    if (pw & 1) (ans *= bas) %= 1000000007;
  return ans;
}
void init() {
  fac[0] = 1;
  for (long long i = 1; i <= 9000; i++)
    fac[i] = 1ll * fac[i - 1] * i % 1000000007;
  for (long long i = 0; i <= 9000; i++)
    inv[i] = 1ll * qpow(fac[i], 1000000007 - 2) % 1000000007;
}
inline long long C(long long m, long long n) {
  return 1ll * fac[m] % 1000000007 * inv[m - n] % 1000000007 * inv[n] %
         1000000007;
}
signed main() {
  init();
  n = read();
  for (long long i = 1; i <= n; i++) a[i] = read(), b[i] = read();
  for (long long i = 1; i <= n; i++) val[-a[i] + 2000][-b[i] + 2000]++;
  for (long long i = 1; i <= 4000; i++) add(val[0][i], val[0][i - 1]);
  for (long long i = 1; i <= 4000; i++) add(val[i][0], val[i - 1][0]);
  for (long long i = 1; i <= 4000; i++)
    for (long long j = 1; j <= 4000; j++)
      add(val[i][j], val[i][j - 1]), add(val[i][j], val[i - 1][j]);
  for (long long i = 1; i <= n; i++) add(ans, val[a[i] + 2000][b[i] + 2000]);
  for (long long i = 1; i <= n; i++)
    add(ans, -C(a[i] + b[i] + a[i] + b[i], a[i] + a[i]));
  cout << ((ans * inv[2]) % 1000000007 + 1000000007) % 1000000007 << endl;
  return 0;
}
