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;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
vector<long long> g;
const long long bas = 10000019, mod = 1000000007, N = 100010;
struct data {
  long long v, w, s, t;
} a[N];
long long n, m, tim, f[40][N], mark[N], cnt, Q;
void work(long long l, long long r, vector<long long> g, long long dep) {
  for (long long i = 1; i <= m; ++i) f[dep][i] = f[dep - 1][i];
  for (long long i = 0; i < g.size(); i++)
    if (a[g[i]].s <= l && a[g[i]].t >= r)
      for (long long j = m; j >= a[g[i]].w; --j)
        f[dep][j] = max(f[dep][j], f[dep][j - a[g[i]].w] + a[g[i]].v);
  if (l == r) {
    if (mark[l]) {
      long long ans = 0;
      for (long long i = m; i >= 1; --i) ans = (ans * bas + f[dep][i]) % mod;
      writeln(ans);
    }
    return;
  } else {
    long long mid = (l + r) >> 1;
    vector<long long> ls, rs;
    for (long long i = 0; i < g.size(); i++) {
      if (a[g[i]].s <= l && a[g[i]].t >= r) continue;
      if (a[g[i]].t <= mid)
        ls.push_back(g[i]);
      else if (a[g[i]].s > mid)
        rs.push_back(g[i]);
      else
        ls.push_back(g[i]), rs.push_back(g[i]);
    }
    work(l, mid, ls, dep + 1);
    work(mid + 1, r, rs, dep + 1);
  }
}
int main() {
  n = read();
  m = read();
  for (long long i = 1; i <= n; ++i)
    a[++cnt].v = read(), a[cnt].w = read(), a[cnt].s = 1;
  Q = read();
  for (long long i = 1; i <= Q; ++i) {
    long long opt = read();
    if (opt == 1) a[++cnt].v = read(), a[cnt].w = read(), a[cnt].s = i;
    if (opt == 2) a[read()].t = i;
    if (opt == 3) mark[i] = 1;
  }
  for (long long i = 1; i <= cnt; ++i)
    if (!a[i].t) a[i].t = Q;
  for (long long i = 1; i <= cnt; ++i) g.push_back(i);
  work(1, Q, g, 1);
}