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 int MaxK = 1010, Log = 17, MaxN = 30010;
const int base = 10000019, mod = 1000000007;
int v[MaxN], w[MaxN], l[MaxN], r[MaxN];
vector<int> T[1 << Log];
int n, k, q, f[MaxK], ans[MaxN], pw[MaxK];
void insert(int l, int r, int now, int x, int y, int z) {
  if (x <= l && r <= y) return T[now].push_back(z);
  if (r < x || y < l) return;
  int mid = (l + r) >> 1;
  insert(l, mid, now << 1, x, y, z);
  insert(mid + 1, r, now << 1 | 1, x, y, z);
}
void solve(int l, int r, int now, int *f) {
  int *g = new int[MaxK];
  memcpy(g, f, (k + 1) << 2);
  for (auto p : T[now]) {
    for (register int i = k; i >= w[p]; --i)
      f[i] = max(f[i], f[i - w[p]] + v[p]);
  }
  if (l == r) {
    for (int i = 1; i <= k; ++i)
      ans[l] = (ans[l] + 1ll * f[i] * pw[i - 1]) % mod;
  } else {
    int mid = (l + r) >> 1;
    solve(l, mid, now << 1, f);
    solve(mid + 1, r, now << 1 | 1, f);
  }
  memcpy(f, g, (k + 1) << 2);
  delete[] g;
}
int main() {
  scanf("%d%d", &n, &k);
  pw[0] = 1;
  for (int i = 1; i <= k; ++i) pw[i] = 1ll * pw[i - 1] * base % mod;
  for (int i = 1; i <= n; ++i) scanf("%d%d", v + i, w + i);
  scanf("%d", &q);
  int now = 0;
  for (int i = 1; i <= n; ++i) l[i] = 1, r[i] = -1;
  for (int i = 1; i <= q; ++i) {
    int type, u;
    scanf("%d", &type);
    if (type == 3)
      ++now;
    else if (type == 2) {
      scanf("%d", &u);
      r[u] = now;
    } else {
      ++n;
      scanf("%d%d", v + n, w + n);
      l[n] = now + 1;
      r[n] = -1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (r[i] == -1) r[i] = now;
    if (l[i] <= r[i]) insert(1, now, 1, l[i], r[i], i);
  }
  solve(1, now, 1, f);
  for (int i = 1; i <= now; ++i) printf("%d\n", ans[i]);
  return 0;
}
