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;
long long tavan(long long x, long long y) {
  long long res = 1;
  while (y) {
    res *= y % 2 ? x : 1;
    res %= ((long long)1e9 + 7);
    x *= x;
    x %= ((long long)1e9 + 7);
    y /= 2;
  }
  return res;
}
struct exhibit {
  long long l, r, v, w;
};
long long n, k, q, num, now[((long long)1020)];
long long ans[((long long)31 * 1000)];
exhibit a[((long long)31 * 1000)];
vector<pair<long long, long long> > seg[4 * ((long long)31 * 1000)];
void add(long long v, long long w, long long dp[]) {
  for (int i = k; i >= w; i--) dp[i] = max(dp[i], dp[i - w] + v);
}
void update(long long ql, long long qr, long long v, long long w,
            long long xl = 0, long long xr = num, long long id = 1) {
  if (xr <= ql || qr <= xl) return;
  if (ql <= xl && xr <= qr) {
    seg[id].push_back({v, w});
    return;
  }
  long long mid = (xl + xr) / 2;
  update(ql, qr, v, w, xl, mid, id * 2);
  update(ql, qr, v, w, mid, xr, id * 2 + 1);
}
void build(long long xl = 0, long long xr = num, long long id = 1) {
  long long ex[((long long)1020)];
  for (int i = 0; i <= k; i++) ex[i] = now[i];
  for (auto u : seg[id]) add(u.first, u.second, now);
  if (xl == xr - 1) {
    for (long long i = 1, ex = 1, ex2 = now[1]; i <= k; i++,
                   ex *= ((long long)1e7 + 19), ex %= ((long long)1e9 + 7),
                   ex2 = max(now[i], ex2))
      if (ex2 > 0) ans[xl] += ex2 * ex, ans[xl] %= ((long long)1e9 + 7);
    for (int i = 0; i <= k; i++) now[i] = ex[i];
    return;
  }
  long long mid = (xl + xr) / 2;
  build(xl, mid, id * 2);
  build(mid, xr, id * 2 + 1);
  for (int i = 0; i <= k; i++) now[i] = ex[i];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(now, -63, sizeof now);
  now[0] = 0;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    long long v, w;
    cin >> v >> w;
    a[i].l = 0, a[i].v = v, a[i].w = w;
    a[i].r = -1;
  }
  cin >> q;
  while (q--) {
    long long t, v, w, x;
    cin >> t;
    if (t == 1) {
      cin >> v >> w;
      a[n].l = num, a[n].v = v, a[n].w = w;
      a[n].r = -1;
      n++;
    }
    if (t == 2) cin >> x, x--, a[x].r = num;
    if (t == 3) num++;
  }
  for (int i = 0; i < n; i++) {
    if (a[i].r == -1) a[i].r = num;
    update(a[i].l, a[i].r, a[i].v, a[i].w);
  }
  build();
  for (int i = 0; i < num; i++) cout << ans[i] << "\n";
  return 0;
}
