Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1, c = getchar();
  while (c < 48) c == '-' && (f = -1), c = getchar();
  while (c > 47) x = x * 10 + c - '0', c = getchar();
  return x * f;
}
const int MAXN = 40005;
const int base = 1e7 + 19, mod = 1e9 + 7;
std::vector<int> adj[MAXN << 2];
int f[25][1005], w[MAXN], v[MAXN], p[MAXN];
int n, m, q, tot, lim;
void modify(int x, int l, int r, int ql, int qr, int k) {
  if (ql > qr) return;
  if (ql <= l && r <= qr) return adj[x].push_back(k), void();
  int mid = (l + r) >> 1;
  if (ql <= mid) modify(x * 2, l, mid, ql, qr, k);
  if (qr > mid) modify(x * 2 + 1, mid + 1, r, ql, qr, k);
}
void solve(int x, int l, int r, int p, int k) {
  if (l > p) return;
  int mid = (l + r) >> 1, res = 0;
  for (int i = 0; i <= m; i++) f[k][i] = f[k - 1][i];
  for (int y : adj[x])
    for (int q = m; q >= w[y]; q--)
      f[k][q] = max(f[k][q], f[k][q - w[y]] + v[y]);
  if (l == r) {
    for (int i = m; i >= 1; i--) res = (1ll * res * base + f[k][i]) % mod;
    return printf("%d\n", res), void();
  }
  solve(x << 1, l, mid, p, k + 1);
  solve(x << 1 | 1, mid + 1, r, p, k + 1);
}
int main(int argc, char const *argv[]) {
  n = read();
  m = read();
  lim = 4e4;
  for (int i = 1; i <= n; ++i) v[i] = read(), w[i] = read(), p[i] = 1;
  for (int q = read(), opt, x; q; --q) {
    opt = read();
    if (opt == 1) v[++n] = read(), w[n] = read(), p[n] = tot + 1;
    if (opt == 2) x = read(), modify(1, 1, lim, p[x], tot, x), p[x] = -1;
    if (opt == 3) ++tot;
  }
  for (int i = 1; i <= n; i++)
    if (~p[i]) modify(1, 1, lim, p[i], tot, i);
  solve(1, 1, lim, tot, 1);
  return 0;
}