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;
void read(int &digit) {
  digit = 0;
  char c;
  for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar())
    ;
  bool type = false;
  if (c == '-') type = true, c = getchar();
  for (; c >= '0' && c <= '9'; digit = digit * 10 + c - '0', c = getchar())
    ;
  if (type == true) digit = -digit;
}
struct Things {
  int w, v;
  int l, r;
} a[35010];
const int base = 1e7 + 19;
const int mo = 1e9 + 7;
int que[20][35010], cnt1[20];
int ques[20][35010], cnt2[20];
int f[20][35010];
int cnt;
int n, m, k;
int mjy[35010];
inline bool in(int l, int r, int ql, int qr) {
  if (l > ql) swap(l, ql), swap(r, qr);
  return (r >= ql);
}
void solve(int l, int r, int dep) {
  if (cnt2[dep] == 0) return;
  for (int i = 1; i <= cnt1[dep]; i++) {
    int u = que[dep][i];
    if (a[u].l <= l && a[u].r >= r) {
      for (int j = k; j >= a[u].w; j--)
        if (f[dep][j] < f[dep][j - a[u].w] + a[u].v)
          f[dep][j] = f[dep][j - a[u].w] + a[u].v;
    }
  }
  if (l == r) {
    int maxx = 0;
    for (int j = 1; j <= k; j++)
      f[dep][j] = max(f[dep][j], maxx), maxx = f[dep][j];
    long long zxo = 0;
    for (int j = k; j >= 1; j--) zxo = (zxo * base + f[dep][j]) % mo;
    for (int i = 1; i <= cnt2[dep]; i++) mjy[ques[dep][i]] = zxo;
    return;
  }
  int mid = (l + r) >> 1;
  cnt1[dep + 1] = cnt2[dep + 1] = 0;
  for (int j = 1; j <= k; j++) f[dep + 1][j] = f[dep][j];
  for (int i = 1; i <= cnt1[dep]; i++) {
    int u = que[dep][i];
    if (!(a[u].l <= l && a[u].r >= r) && in(a[u].l, a[u].r, l, mid))
      que[dep + 1][++cnt1[dep + 1]] = u;
  }
  for (int i = 1; i <= cnt2[dep]; i++)
    if (ques[dep][i] <= mid) {
      ques[dep + 1][++cnt2[dep + 1]] = ques[dep][i];
    }
  solve(l, mid, dep + 1);
  cnt1[dep + 1] = cnt2[dep + 1] = 0;
  for (int j = 1; j <= k; j++) f[dep + 1][j] = f[dep][j];
  for (int i = 1; i <= cnt1[dep]; i++) {
    int u = que[dep][i];
    if (!(a[u].l <= l && a[u].r >= r) && in(a[u].l, a[u].r, mid + 1, r))
      que[dep + 1][++cnt1[dep + 1]] = u;
  }
  for (int i = 1; i <= cnt2[dep]; i++)
    if (ques[dep][i] > mid) {
      ques[dep + 1][++cnt2[dep + 1]] = ques[dep][i];
    }
  solve(mid + 1, r, dep + 1);
}
int main() {
  read(n), read(k);
  for (int i = 1; i <= n; i++) read(a[i].v), read(a[i].w);
  read(m);
  for (int i = 1; i <= n; i++) a[i].l = 1, a[i].r = m;
  cnt = n;
  for (int i = 1; i <= m; i++) {
    int op;
    read(op);
    if (op == 1) {
      cnt++;
      read(a[cnt].v), read(a[cnt].w);
      a[cnt].l = i;
      a[cnt].r = m;
    }
    if (op == 2) {
      int x;
      read(x);
      a[x].r = i;
    }
    if (op == 3) {
      ques[0][++cnt2[0]] = i;
    }
  }
  for (int i = 1; i <= m; i++) mjy[i] = -1;
  for (int i = 1; i <= cnt; i++) que[0][++cnt1[0]] = i;
  solve(1, m, 0);
  for (int i = 1; i <= m; i++)
    if (mjy[i] != -1) printf("%d\n", mjy[i]);
  return 0;
}
