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;
template <typename T>
inline bool chkmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
const int oo = 0x3f3f3f3f;
const long long OO = 0x3f3f3f3f3f3f3f3fLL;
template <typename T>
inline T &Read(T &first) {
  static char c;
  while (1) {
    c = getchar();
    if (c == '-' || (c >= '0' && c <= '9')) break;
  }
  bool flag = c == '-';
  first = flag ? 0 : c - '0';
  while (1) {
    c = getchar();
    if (c < '0' || c > '9') break;
    (first *= 10) += c - '0';
  }
  if (flag) first = -first;
  return first;
}
const int maxn = 15000, maxm = 1000, maxq = 35000, max0 = 16;
const int p = 1e7 + 19, q = 1e9 + 7;
int n, K;
int opr[maxq + 5];
int v[maxn + 5], w[maxn + 5];
int ans[max0 + 1][maxm + 5];
int del[maxn + 5];
bool in[maxn + 5];
inline void add(int *ans, int first) {
  for (int i = K - w[first]; i >= 0; --i)
    chkmax(ans[i + w[first]], ans[i] + v[first]);
}
inline void work(int l, int r, int dep) {
  bool flag = 0;
  for (int i = (l), _end_ = (r); i < _end_; ++i)
    if (opr[i] == -1) {
      flag = 1;
      break;
    }
  if (!flag) {
    for (int i = (l), _end_ = (r); i < _end_; ++i)
      if (!(opr[i] & 1)) in[opr[i] >> 1] = 1;
    return;
  }
  if (r - l <= 1) {
    long long ret = 0, t = 1;
    for (int i = (1), _end_ = (K + 1); i < _end_; ++i) {
      (ret += t * ans[dep][i]) %= q;
      (t *= p) %= q;
    }
    printf("%I64d\n", ret);
    return;
  }
  memcpy(ans[dep + 1], ans[dep], sizeof ans[dep + 1]);
  int mid = (l + r) >> 1;
  for (int i = (mid), _end_ = (r); i < _end_; ++i) {
    if (opr[i] >= 0) {
      if ((opr[i] & 1) && in[opr[i] >> 1]) add(ans[dep + 1], opr[i] >> 1);
    }
  }
  work(l, mid, dep + 1);
  memcpy(ans[dep + 1], ans[dep], sizeof ans[dep + 1]);
  for (int i = (l), _end_ = (mid); i < _end_; ++i) {
    if (opr[i] >= 0) {
      if (!(opr[i] & 1) && del[opr[i] >> 1] >= r)
        add(ans[dep + 1], opr[i] >> 1);
    }
  }
  work(mid, r, dep + 1);
}
int main() {
  srand(time(NULL));
  scanf("%d%d", &n, &K);
  int qn;
  memset(del, oo, sizeof del);
  for (int i = (0), _end_ = (n); i < _end_; ++i)
    scanf("%d%d", v + i, w + i), opr[i] = i << 1;
  scanf("%d", &qn);
  int tmp = qn + n;
  for (int i = (n), _end_ = (qn + n); i < _end_; ++i) {
    int ty;
    scanf("%d", &ty);
    if (ty == 1) {
      scanf("%d%d", v + n, w + n);
      opr[i] = n << 1;
      ++n;
    } else if (ty == 2) {
      scanf("%d", opr + i), --opr[i];
      del[opr[i]] = i;
      (opr[i] <<= 1) |= 1;
    } else
      opr[i] = -1;
  }
  qn = tmp;
  work(0, qn, 0);
  return 0;
}
