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 long long MOD = 1000000007LL;
const long long pra = 10000019LL;
int v[15000 + 10];
int w[15000 + 10];
int zac[15000 + 10];
int kon[15000 + 10];
int q[2 * 15000 + 10];
long long memo[15000 + 10][1010];
long long pwrs[1010];
vector<int> stree[8 * 15000 + 10];
int n, k;
void add(int poz, int l, int d, int el) {
  if (zac[el] <= l && kon[el] >= d) {
    stree[poz].push_back(el);
    return;
  }
  if (zac[el] > d || kon[el] < l) return;
  int s = (l + d) / 2;
  add(poz * 2, l, s, el);
  add(poz * 2 + 1, s + 1, d, el);
}
void solve(int poz, int l, int d, int done) {
  for (int i = 0, _n = stree[poz].size(); i < _n; i++) {
    int kje = done + 1 + i;
    int predmet = stree[poz][i];
    for (int j = 0, _n = k + 1; j < _n; j++) {
      if (j - w[predmet] >= 0)
        memo[kje][j] = memo[kje - 1][j - w[predmet]] + v[predmet];
      else
        memo[kje][j] = 0;
      if (memo[kje][j - 1] > memo[kje][j]) memo[kje][j] = memo[kje][j - 1];
      if (memo[kje - 1][j] > memo[kje][j]) memo[kje][j] = memo[kje - 1][j];
    }
  }
  int skupaj = done + stree[poz].size();
  if (l == d) {
    long long rez = 0LL;
    for (int j = 1, _n = k + 1; j < _n; j++)
      rez = (rez + memo[skupaj][j] * pwrs[j - 1]) % MOD;
    printf("%d\n", (int)(rez));
    return;
  }
  int s = (l + d) / 2;
  solve(poz * 2, l, s, skupaj);
  solve(poz * 2 + 1, s + 1, d, skupaj);
  return;
}
int main() {
  int stq = 0;
  scanf("%d%d", &n, &k);
  for (int i = 0, _n = n; i < _n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    v[i] = a;
    w[i] = b;
    zac[i] = 0;
    kon[i] = -2;
  }
  int nq;
  scanf("%d", &nq);
  for (int i = 0, _n = nq; i < _n; i++) {
    int st;
    scanf("%d", &st);
    if (st == 3) {
      q[stq++] = i;
    }
    if (st == 2) {
      int gg;
      scanf("%d", &gg);
      kon[--gg] = stq - 1;
    }
    if (st == 1) {
      int a, b;
      scanf("%d%d", &a, &b);
      v[n] = a;
      w[n] = b;
      zac[n] = stq;
      kon[n++] = -2;
    }
  }
  for (int i = 0, _n = n; i < _n; i++)
    if (kon[i] == -2) kon[i] = stq - 1;
  for (int i = 0, _n = n; i < _n; i++)
    if (kon[i] >= zac[i]) add(1, 0, stq - 1, i);
  pwrs[0] = 1;
  for (int i = 1, _n = 1003; i < _n; i++) pwrs[i] = (pra * pwrs[i - 1]) % MOD;
  solve(1, 0, stq - 1, 0);
  return 0;
}
