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;
const int MN = 30111;
const long long MOD = 1e9 + 7;
const long long P = 1e7 + 19;
struct Obj {
  int w, v, from, to;
} a[MN];
int res[MN];
long long f[22][MN], p[MN];
vector<int> it[MN * 4];
int nQuery, nObj, k;
void init(int i, int l, int r) {
  it[i].clear();
  if (l == r) return;
  int mid = (l + r) >> 1;
  init(i << 1, l, mid);
  init(i << 1 | 1, mid + 1, r);
}
void add(int i, int l, int r, int x, int u, int v) {
  if (v < l || r < u) return;
  if (u <= l && r <= v) {
    it[i].push_back(x);
    return;
  }
  int mid = (l + r) >> 1;
  add(i << 1, l, mid, x, u, v);
  add(i << 1 | 1, mid + 1, r, x, u, v);
}
void go(int i, int l, int r, int h) {
  for (int i = (0), _b = (k); i <= _b; ++i) f[h][i] = f[h - 1][i];
  for (int x : it[i]) {
    int w = a[x].w;
    int v = a[x].v;
    for (int sumw = (k), _b = (w); sumw >= _b; --sumw)
      f[h][sumw] = max(f[h][sumw], f[h][sumw - w] + v);
  }
  if (l == r) {
    res[l] = 0;
    for (int m = (1), _b = (k); m <= _b; ++m)
      res[l] = (res[l] + f[h][m] * p[m - 1]) % MOD;
    return;
  }
  int mid = (l + r) >> 1;
  go(i << 1, l, mid, h + 1);
  go(i << 1 | 1, mid + 1, r, h + 1);
}
int main() {
  int n, q;
  p[0] = 1;
  for (int i = (1), _b = (MN - 1); i <= _b; ++i) p[i] = p[i - 1] * P % MOD;
  while (scanf("%d%d", &n, &k) == 2) {
    nObj = 0;
    for (int i = (1), _b = (n); i <= _b; ++i) {
      int v, w;
      scanf("%d%d", &v, &w);
      ++nObj;
      a[nObj].w = w;
      a[nObj].v = v;
      a[nObj].from = 1;
      a[nObj].to = -1;
    }
    nQuery = 0;
    scanf("%d", &q);
    for (int i = (1), _b = (q); i <= _b; ++i) {
      int typ;
      scanf("%d", &typ);
      if (typ == 1) {
        ++nObj;
        scanf("%d%d", &a[nObj].v, &a[nObj].w);
        a[nObj].from = nQuery + 1;
        a[nObj].to = -1;
      } else if (typ == 2) {
        int x;
        scanf("%d", &x);
        a[x].to = nQuery;
      } else {
        ++nQuery;
      }
    }
    for (int i = (1), _b = (nObj); i <= _b; ++i)
      if (a[i].to < 0) a[i].to = nQuery;
    init(1, 1, nQuery);
    for (int i = (1), _b = (nObj); i <= _b; ++i) {
      add(1, 1, nQuery, i, a[i].from, a[i].to);
    }
    go(1, 1, nQuery, 1);
    for (int i = (1), _b = (nQuery); i <= _b; ++i) printf("%d\n", (int)res[i]);
    puts("");
  }
}