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 MN = 15005, MK = 1005, MQ = 30005, inf = 1000000005, P2 = (1 << 15),
          MTS = 2 * P2 + 5;
const long long INF = 1000000000000000005LL, mod = (long long)1e9 + 7LL,
                p = (long long)1e7 + 19LL;
long long dp[MN][MK];
vector<int> T[MTS];
long long ans[MTS], pot[MK];
int mass[MN], val[MN], pocz[MN], kon[MN], chce[MQ];
int k, idx;
long long daj() {
  long long ret = 0LL;
  for (int i = 1; i <= k; ++i) {
    ret += dp[idx][i] * pot[i - 1];
    ret %= mod;
  }
  return ret;
}
void dodaj(int x) {
  idx++;
  int v = val[x], m = mass[x];
  for (int i = 1; i < m; ++i) dp[idx][i] = dp[idx - 1][i];
  for (int i = m; i <= k; ++i)
    dp[idx][i] =
        max(dp[idx][i - 1], max(dp[idx - 1][i], v + dp[idx - 1][i - m]));
}
void wrzuc(int a, int b, int x) {
  a += P2;
  b += P2;
  while (a < b) {
    if (a & 1) T[a++].push_back(x);
    if (!(b & 1)) T[b--].push_back(x);
    a >>= 1, b >>= 1;
  }
  if (a == b) T[a].push_back(x);
}
void solve(int x) {
  for (auto v : T[x]) dodaj(v);
  if (x < P2) {
    solve(x << 1);
    solve(1 + (x << 1));
  } else {
    ans[x - P2] = daj();
  }
  idx -= T[x].size();
}
int main() {
  int n;
  scanf("%d%d", &n, &k);
  pot[0] = 1LL;
  for (int i = 1; i <= k; ++i) pot[i] = (pot[i - 1] * p) % mod;
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", &val[i], &mass[i]);
    pocz[i] = 1;
  }
  int ost = n;
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    int typ;
    scanf("%d", &typ);
    if (typ == 1) {
      pocz[++ost] = i;
      scanf("%d%d", &val[ost], &mass[ost]);
    }
    if (typ == 2) {
      int x;
      scanf("%d", &x);
      kon[x] = i;
    }
    if (typ == 3) {
      chce[i] = 1;
    }
  }
  for (int i = 1; i <= ost; ++i) {
    if (!kon[i]) kon[i] = q;
    wrzuc(pocz[i], kon[i], i);
  }
  solve(1);
  for (int i = 1; i <= q; ++i)
    if (chce[i]) printf("%lld\n", ans[i]);
}
