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 N = 30030, M = 5005000, L = 1010, oo = 1000000000, P = 10000019,
          Q = 1000000007;
int i, j, k, n, m, nm, Tn, q, En, ch, last, o, x, An;
int f[22][L], h[N << 2], ans[N], p[N];
struct cc {
  int l, r, v, w;
} A[N];
struct edge {
  int s, n;
} E[M];
void R(int &x) {
  x = 0;
  ch = getchar();
  while (ch < '0' || '9' < ch) ch = getchar();
  while ('0' <= ch && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
}
int max(const int &x, const int &y) {
  if (x > y) return x;
  return y;
}
void E_add(int x, int y) {
  E[++En].s = y;
  E[En].n = h[x];
  h[x] = En;
}
void T_add(int L, int R, int l, int r, int x, int k) {
  if (L == l && R == r) {
    E_add(k, x);
    return;
  }
  int mid = (L + R) >> 1;
  if (r <= mid)
    T_add(L, mid, l, r, x, k << 1);
  else {
    if (l > mid)
      T_add(mid + 1, R, l, r, x, k << 1 | 1);
    else
      T_add(L, mid, l, mid, x, k << 1),
          T_add(mid + 1, R, mid + 1, r, x, k << 1 | 1);
  }
}
void T_bl(int L, int R, int dp, int k) {
  for (int i = 0; i <= m; i++) f[dp][i] = f[dp - 1][i];
  for (int kk = h[k]; kk; kk = E[kk].n) {
    int v = A[E[kk].s].v, w = A[E[kk].s].w;
    for (int i = m; i >= w; i--) f[dp][i] = max(f[dp][i], f[dp][i - w] + v);
  }
  if (L == R) {
    int &res = ans[L];
    int pp = 1;
    for (int i = 1; i <= m; i++) {
      res = ((long long)f[dp][i] * pp + res) % Q;
      pp = (long long)pp * P % Q;
    }
    return;
  }
  int mid = (L + R) >> 1;
  T_bl(L, mid, dp + 1, k << 1);
  T_bl(mid + 1, R, dp + 1, k << 1 | 1);
}
int main() {
  R(n);
  R(m);
  for (i = 1; i <= n; i++) {
    A[i].l = 1;
    A[i].r = oo;
    R(A[i].v);
    R(A[i].w);
  }
  R(q);
  An = n;
  for (i = 1; i <= q; i++) {
    last = o;
    R(o);
    if (o == 3) {
      if (last == 3)
        p[++nm] = Tn;
      else
        p[++nm] = ++Tn;
      continue;
    }
    if (o == 1) {
      An++;
      A[An].l = Tn + 1;
      A[An].r = oo;
      R(A[An].v);
      R(A[An].w);
      continue;
    }
    if (o == 2) {
      R(x);
      A[x].r = Tn;
      continue;
    }
  }
  for (i = 1; i <= An; i++) {
    if (A[i].r == oo) A[i].r = Tn;
    if (1 <= A[i].l && A[i].l <= A[i].r && A[i].r <= Tn)
      T_add(1, Tn, A[i].l, A[i].r, i, 1);
  }
  T_bl(1, Tn, 1, 1);
  for (i = 1; i <= nm; i++) printf("%d\n", ans[p[i]]);
}