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>
struct edge {
  int v, w;
  edge *nt;
} t[65536 << 3], *e[65536], *V = t;
int v[35001], w[35001], k, f[1001 * 17], p[1001], le[35001], wt, ss[19];
char fl[1 << 23], *A = fl;
bool u[65536];
void read(int &x) {
  char c;
  for (x = 0; '0' > (c = *A++) || c > '9';)
    ;
  while ('0' <= c && c <= '9') x = (x << 3) + (x << 1) + c - 48, (c = *A++);
}
void print(int x) {
  if (!x)
    *A++ = (48);
  else {
    for (wt = 0; x; ss[++wt] = x % 10, x /= 10)
      ;
    for (; wt; *A++ = (ss[wt--] + 48))
      ;
  }
  *A++ = '\n';
}
void dfs(int x, int l, int r, int f[]) {
  if (!u[x]) return;
  for (edge *v = e[x]; v; v = v->nt)
    for (int i = k; i >= v->w; i--)
      if (f[i] < f[i - v->w] + v->v) f[i] = f[i - v->w] + v->v;
  if (l == r) {
    int MAPLE = 0;
    for (int i = 1; i <= k;
         (MAPLE += (long long)f[i] * p[i] % 1000000007) %= 1000000007, i++)
      ;
    print(MAPLE);
    return;
  }
  int *g = f + 1001, mid = (l + r) >> 1;
  for (int i = 1; i <= k; g[i] = f[i], i++)
    ;
  dfs(x << 1, l, mid, g);
  for (int i = 1; i <= k; g[i] = f[i], i++)
    ;
  dfs(x << 1 | 1, mid + 1, r, g);
}
int main() {
  int n, q, i, l, r, x, y, M;
  for (*(fl + fread(fl, 1, 1 << 23, stdin)) = EOF, read(n), read(k), i = 1;
       i <= n; read(v[i]), read(w[i]), le[i] = 1, i++)
    ;
  for (read(q), M = 1; M <= q + 1; M <<= 1)
    ;
  for (p[1] = 1, i = 2; i <= k;
       p[i] = (long long)p[i - 1] * 10000019 % 1000000007, i++)
    ;
  for (i = 1; i <= q; i++) switch (read(r), r) {
      case 1:
        read(v[++n]), read(w[n]), le[n] = i;
        break;
      case 2:
        for (read(r), x = v[r], y = w[r], l = le[r] + M - 1, le[r] = 10000019,
                      r = i + M + 1;
             l ^ r ^ 1; l >>= 1, r >>= 1) {
          if (~l & 1) *(++V) = (edge){x, y, e[l ^ 1]}, e[l ^ 1] = V;
          if (r & 1) *(++V) = (edge){x, y, e[r ^ 1]}, e[r ^ 1] = V;
        }
        break;
      case 3:
        for (l = i + M; l; u[l] = true, l >>= 1)
          ;
        break;
    }
  for (i = 1; i <= n; i++)
    if (le[i] != 10000019)
      for (x = v[i], y = w[i], l = le[i] + M - 1, r = q + M + 1; l ^ r ^ 1;
           l >>= 1, r >>= 1) {
        if (~l & 1) *(++V) = (edge){x, y, e[l ^ 1]}, e[l ^ 1] = V;
        if (r & 1) *(++V) = (edge){x, y, e[r ^ 1]}, e[r ^ 1] = V;
      }
  A = fl, dfs(1, 0, M - 1, f), fwrite(fl, 1, A - fl, stdout);
}