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;
template <class T, class L>
bool smax(T &x, L y) {
  return x < y ? (x = y, 1) : 0;
}
template <class T, class L>
bool smin(T &x, L y) {
  return x > y ? (x = y, 1) : 0;
}
const int maxn = 15e3 + 17, maxk = 1e3 + 17, mod = 1e9 + 7, base = 1e7 + 19,
          lg = 17, maxq = 3e4 + 17;
vector<int> iman[maxq << 2];
int n, k, nq, q, dp[lg][maxk], w[maxn], v[maxn], ans[maxq], po[maxk];
void majid(int st, int en, int idx, int l = 0, int r = q, int id = 1) {
  if (st <= l && r <= en) {
    iman[id].push_back(idx);
    return;
  }
  if (en <= l || r <= st) return;
  int mid = l + r >> 1;
  majid(st, en, idx, l, mid, id << 1), majid(st, en, idx, mid, r, id << 1 | 1);
}
void hamid(int l = 0, int r = q, int id = 1, int h = 1) {
  memcpy(dp[h], dp[h - 1], sizeof dp[h]);
  for (auto idx : iman[id])
    for (int wi = k - w[idx]; wi >= 0; wi--)
      smax(dp[h][wi + w[idx]], dp[h][wi] + v[idx]);
  partial_sum(dp[h], dp[h] + k + 1, dp[h],
              [](int &x, int &y) { return max(x, y); });
  if (r - l < 2) {
    for (int i = 1; i <= k; i++)
      ans[l] = (ans[l] + (long long)po[i - 1] * dp[h][i]) % mod;
    return;
  }
  int mid = l + r >> 1;
  hamid(l, mid, id << 1, h + 1), hamid(mid, r, id << 1 | 1, h + 1);
}
int st[maxn], en[maxn];
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> v[i] >> w[i];
  po[0] = 1;
  for (int i = 1; i <= k; i++) po[i] = (long long)po[i - 1] * base % mod;
  memset(en, 63, sizeof en);
  cin >> nq;
  for (int t, x; nq--;) {
    cin >> t;
    if (t == 1)
      cin >> v[n] >> w[n], st[n++] = q;
    else if (t == 2)
      cin >> x, en[x - 1] = q;
    else
      q++;
  }
  for (int i = 0; i < n; i++) majid(st[i], min(q, en[i]), i);
  hamid();
  for (int i = 0; i < q; i++) cout << ans[i] << '\n';
  return 0;
}
