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;
vector<int> tag[100100 << 2];
long long tmp[20][1010];
int n, ans[100100], k, m, q, val[100100], cost[100100], qes[100100], st[100100],
    ed[100100];
void update(int rt, int l, int r, int L, int R, int id) {
  if (L > R) return;
  if (l >= L && r <= R) {
    tag[rt].push_back(id);
    return;
  }
  if (L <= ((l + r) >> 1)) update(rt << 1, l, ((l + r) >> 1), L, R, id);
  if (R > ((l + r) >> 1)) update(rt << 1 | 1, ((l + r) >> 1) + 1, r, L, R, id);
}
void Max(long long &f, long long g) {
  if (f < g) f = g;
}
void dfs(int rt, int l, int r, int dep) {
  memcpy(tmp[dep], tmp[dep - 1], sizeof tmp[dep]);
  int sz = tag[rt].size();
  for (int i = 0; i <= sz - 1; i++) {
    int id = tag[rt][i];
    for (int j = k; j >= cost[id]; j--)
      Max(tmp[dep][j], tmp[dep][j - cost[id]] + val[id]);
  }
  if (l == r) {
    for (int i = k; i >= 1; i--)
      ans[l] = (1LL * ans[l] * 10000019 + tmp[dep][i]) % 1000000007;
    return;
  }
  dfs(rt << 1, l, ((l + r) >> 1), dep + 1);
  dfs(rt << 1 | 1, ((l + r) >> 1) + 1, r, dep + 1);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d%d", val + i, cost + i), st[i] = 0;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int opt, x;
    scanf("%d", &opt);
    if (opt == 1) ++n, scanf("%d%d", val + n, cost + n), st[n] = i;
    if (opt == 2) scanf("%d", &x), ed[x] = i;
    if (opt == 3) qes[++*qes] = i;
  }
  for (int i = 1; i <= n; i++) {
    st[i] = upper_bound(qes + 1, qes + *qes + 1, st[i]) - qes;
    if (ed[i])
      ed[i] = upper_bound(qes + 1, qes + *qes + 1, ed[i]) - qes - 1;
    else
      ed[i] = *qes;
  }
  for (int i = 1; i <= n; i++) update(1, 1, *qes, st[i], ed[i], i);
  dfs(1, 1, *qes, 1);
  for (int i = 1; i <= *qes; i++) printf("%d\n", ans[i]);
}