There's a famous museum in the city where Kleofáš lives. In the museum, n exhibits (numbered 1 through n) had been displayed for a long time; the i-th of those exhibits has value vi and mass wi. 

Then, the museum was bought by a large financial group and started to vary the exhibits. At about the same time, Kleofáš... gained interest in the museum, so to say.

You should process q events of three types:

  * type 1 — the museum displays an exhibit with value v and mass w; the exhibit displayed in the i-th event of this type is numbered n + i (see sample explanation for more details)
  * type 2 — the museum removes the exhibit with number x and stores it safely in its vault
  * type 3 — Kleofáš visits the museum and wonders (for no important reason at all, of course): if there was a robbery and exhibits with total mass at most m were stolen, what would their maximum possible total value be?



For each event of type 3, let s(m) be the maximum possible total value of stolen exhibits with total mass  ≤ m.

Formally, let D be the set of numbers of all exhibits that are currently displayed (so initially D = {1, ..., n}). Let P(D) be the set of all subsets of D and let 

<image>

Then, s(m) is defined as 

<image>

Compute s(m) for each <image>. Note that the output follows a special format.

Input

The first line of the input contains two space-separated integers n and k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 1000) — the initial number of exhibits in the museum and the maximum interesting mass of stolen exhibits. 

Then, n lines follow. The i-th of them contains two space-separated positive integers vi and wi (1 ≤ vi ≤ 1 000 000, 1 ≤ wi ≤ 1000) — the value and mass of the i-th exhibit.

The next line contains a single integer q (1 ≤ q ≤ 30 000) — the number of events.

Each of the next q lines contains the description of one event in the following format:

  * 1 v w — an event of type 1, a new exhibit with value v and mass w has been added (1 ≤ v ≤ 1 000 000, 1 ≤ w ≤ 1000)
  * 2 x — an event of type 2, the exhibit with number x has been removed; it's guaranteed that the removed exhibit had been displayed at that time
  * 3 — an event of type 3, Kleofáš visits the museum and asks his question



There will be at most 10 000 events of type 1 and at least one event of type 3.

Output

As the number of values s(m) can get large, output the answers to events of type 3 in a special format.

For each event of type 3, consider the values s(m) computed for the question that Kleofáš asked in this event; print one line containing a single number 

<image>

where p = 107 + 19 and q = 109 + 7.

Print the answers to events of type 3 in the order in which they appear in the input.

Examples

Input

3 10
30 4
60 6
5 1
9
3
1 42 5
1 20 3
3
2 2
2 4
3
1 40 6
3


Output

556674384
168191145
947033915
181541912


Input

3 1000
100 42
100 47
400 15
4
2 2
2 1
2 3
3


Output

0

Note

In the first sample, the numbers of displayed exhibits and values s(1), ..., s(10) for individual events of type 3 are, in order: 

<image> <image> <image> <image>

The values of individual exhibits are v1 = 30, v2 = 60, v3 = 5, v4 = 42, v5 = 20, v6 = 40 and their masses are w1 = 4, w2 = 6, w3 = 1, w4 = 5, w5 = 3, w6 = 6.

In the second sample, the only question is asked after removing all exhibits, so s(m) = 0 for any m.
#include <bits/stdc++.h>
using namespace std;
const long long p = 1e7 + 19, q = 1e9 + 7;
long long n, m, k, f[1005], pw[1005], pos[35005], pos1[35005], tot;
pair<long long, long long> e[35005];
bool qr[30005];
vector<pair<long long, long long> > g[120005];
long long read() {
  long long f = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    f = f * 10 + ch - '0';
    ch = getchar();
  }
  return f * w;
}
void modify(long long tr, long long stdl, long long stdr, long long l,
            long long r, long long x, long long y) {
  if (l > stdr || r < stdl) return;
  if (l <= stdl && r >= stdr) {
    g[tr].push_back(make_pair(x, y));
    return;
  }
  long long mid = (stdl + stdr) / 2;
  modify(tr * 2, stdl, mid, l, r, x, y),
      modify(tr * 2 + 1, mid + 1, stdr, l, r, x, y);
}
void solve(long long tr, long long l, long long r) {
  long long tmp[1005] = {0};
  for (long long i = 0; i < g[tr].size(); i++)
    for (long long j = k; j >= 0; j--)
      if (j + g[tr][i].first <= k)
        f[j + g[tr][i].first] =
            max(f[j + g[tr][i].first], f[j] + g[tr][i].second);
  if (l == r) {
    if (qr[l]) {
      long long ans = 0;
      for (long long i = 1; i <= k; i++) ans = (ans + f[i] * pw[i - 1] % q) % q;
      printf("%lld\n", ans);
    }
    return;
  }
  long long mid = (l + r) / 2;
  memcpy(tmp, f, sizeof(tmp));
  solve(tr * 2, l, mid);
  memcpy(f, tmp, sizeof(f));
  solve(tr * 2 + 1, mid + 1, r);
}
signed main() {
  n = read(), k = read();
  pw[0] = 1;
  for (long long i = 1; i <= 1000; i++) pw[i] = pw[i - 1] * p % q;
  for (long long i = 1; i <= n; i++)
    e[i].second = read(), e[i].first = read(), pos[++tot] = 1;
  m = read();
  for (long long i = 1; i <= m; i++) {
    long long opt = read();
    if (opt == 1) {
      long long x = read(), y = read();
      e[++tot] = make_pair(y, x), pos[tot] = i;
    } else if (opt == 2) {
      long long x = read();
      pos1[x] = i;
    } else
      qr[i] = 1;
  }
  for (long long i = 1; i <= tot; i++) {
    if (!pos1[i])
      modify(1, 1, m, pos[i], q, e[i].first, e[i].second);
    else
      modify(1, 1, m, pos[i], pos1[i], e[i].first, e[i].second);
  }
  solve(1, 1, m);
  return 0;
}
