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 iinf = 1e9 + 7;
const long long linf = 1ll << 60;
const double dinf = 1e10;
template <typename T>
inline void scf(T &x) {
  bool f = 0;
  x = 0;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') {
    f = 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  if (f) x = -x;
  return;
}
template <typename T1, typename T2>
void scf(T1 &x, T2 &y) {
  scf(x);
  return scf(y);
}
template <typename T1, typename T2, typename T3>
void scf(T1 &x, T2 &y, T3 &z) {
  scf(x);
  scf(y);
  return scf(z);
}
const int N = 2e4 + 100;
const int K = 1e3 + 10;
const long long _p = 10000019ll;
const long long _q = 1000000007ll;
int n, k, q, tim;
int dp[K];
int tin[N], tout[N];
int v[N], w[N];
long long _pow[N];
bool used[N];
inline int MAX(const int &x, const int &y) { return x < y ? y : x; }
inline void solve(int l = 1, int r = tim) {
  int pd[K], q[N];
  int rar = 0;
  memcpy(pd, dp, sizeof(dp));
  for (int i = (1); i <= (n); i++)
    if (tin[i] < l && tout[i] >= r && !used[i]) {
      used[i] = 1;
      q[rar++] = i;
      for (int j = (k); j >= (w[i]); j--)
        dp[j] = MAX(dp[j], dp[j - w[i]] + v[i]);
    }
  if (l == r) {
    long long cur = 0;
    for (int i = (1); i <= (k); i++)
      cur = (cur + _pow[i - 1] * (long long)dp[i]) % _q;
    printf("%I64d\n", cur);
  } else {
    int mid = l + r >> 1;
    solve(l, mid);
    solve(mid + 1, r);
  }
  for (int i = 0; i < (rar); i++) used[q[i]] = 0;
  memcpy(dp, pd, sizeof(pd));
  return;
}
int main() {
  _pow[0] = 1ll;
  scf(n, k);
  for (int i = (1); i <= (k); i++) _pow[i] = _pow[i - 1] * _p % _q;
  for (int i = (1); i <= (n); i++) scf(v[i], w[i]), tout[i] = -1;
  scf(q);
  for (int i = (1); i <= (q); i++) {
    int op;
    scf(op);
    if (op == 1) {
      n++;
      scf(v[n], w[n]);
      tin[n] = tim;
      tout[n] = -1;
    } else if (op == 2) {
      int x;
      scf(x);
      tout[x] = tim;
    } else
      tim++;
  }
  for (int i = (1); i <= (n); i++)
    if (tout[i] < 0) tout[i] = tim;
  solve();
  return 0;
}
