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;
int readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
struct node {
  int st, ed, v, w;
  node(int ss = 0, int ee = 0, int vv = 0, int ww = 0)
      : st(ss), ed(ee), v(vv), w(ww) {}
} a[50005];
const int cys = 1000000007, jkp = 10000019;
int n, m, k, top, ttm = 1;
int d[20][1005], pw[1005];
vector<int> occ[100005];
int mod(int x) { return x >= cys ? x - cys : x; }
void add(int pl, int x) {
  for (int i = k; i >= a[x].w; i--)
    d[pl][i] = max(d[pl][i], d[pl][i - a[x].w] + a[x].v);
}
void solve(int id, int l, int r) {
  if (l == r) {
    top++;
    for (int i = 1; i <= k; i++) d[top][i] = d[top - 1][i];
    for (auto x : occ[id]) add(top, x);
    int ans = 0;
    for (int i = 1; i <= k; i++)
      ans = mod(ans + 1ll * pw[i - 1] * d[top][i] % cys);
    printf("%d\n", ans);
    top--;
    return;
  }
  int mid = (l + r) / 2;
  top++;
  for (int i = 1; i <= k; i++) d[top][i] = d[top - 1][i];
  for (auto x : occ[id]) add(top, x);
  solve(id << 1, l, mid);
  solve(id << 1 | 1, mid + 1, r);
  top--;
}
void ins(int id, int l, int r, int ql, int qr, int c) {
  if (ql > qr) return;
  if (l == ql && r == qr) {
    occ[id].push_back(c);
    return;
  }
  int mid = (l + r) / 2;
  if (qr <= mid)
    ins(id << 1, l, mid, ql, qr, c);
  else if (ql > mid)
    ins(id << 1 | 1, mid + 1, r, ql, qr, c);
  else
    ins(id << 1, l, mid, ql, mid, c),
        ins(id << 1 | 1, mid + 1, r, mid + 1, qr, c);
}
int main() {
  n = readint();
  k = readint();
  pw[0] = 1;
  for (int i = 1; i <= k; i++) pw[i] = 1ll * pw[i - 1] * jkp % cys;
  int opt, x, y;
  for (int i = 1; i <= n; i++) {
    x = readint();
    y = readint();
    a[i] = node(1, -1, x, y);
  }
  m = readint();
  for (int i = 1; i <= m; i++) {
    opt = readint();
    if (opt == 1) {
      x = readint();
      y = readint();
      a[++n] = node(ttm, -1, x, y);
    } else if (opt == 2)
      a[readint()].ed = ttm - 1;
    else
      ttm++;
  }
  for (int i = 1; i <= n; i++)
    if (a[i].ed < 0) a[i].ed = ttm;
  for (int i = 1; i <= n; i++)
    if (a[i].st <= ttm - 1)
      ins(1, 1, ttm - 1, a[i].st, min(a[i].ed, ttm - 1), i);
  solve(1, 1, ttm - 1);
  return 0;
}
