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>
const int N = (5000 + 30000 << 1) + 5;
const int P = 1e7 + 19;
const int Q = 1e9 + 7;
int V[N], W[N], st[N], ed[N];
int n, m, nq;
int op[N];
int dp[20][1000 + 1];
int answer[N];
int idx[N];
int get_id(int l, int r) { return l + r | l != r; }
std::vector<int> t[N << 1];
void insert(int L, int R, int val, int l, int r) {
  if (L <= l && r <= R) {
    t[get_id(l, r)].push_back(val);
    return;
  }
  int mid = l + r >> 1;
  if (L <= mid) insert(L, R, val, l, mid);
  if (mid < R) insert(L, R, val, mid + 1, r);
}
void solve(int l, int r, int at) {
  memcpy(dp[at], dp[at - 1], sizeof(dp[at - 1]));
  for (auto x : t[get_id(l, r)]) {
    for (int i = m; i >= W[x]; --i) {
      dp[at][i] = std::max(dp[at][i], dp[at][i - W[x]] + V[x]);
    }
  }
  if (l == r) {
    if (op[l] == 2) {
      int val = 0;
      for (int i = 1, slop = 1; i <= m; ++i) {
        (val += dp[at][i] * 1ll * slop % Q) %= Q;
        slop = slop * 1ll * P % Q;
      }
      answer[l] = val;
    }
    return;
  }
  int mid = l + r >> 1;
  solve(l, mid, at + 1);
  solve(mid + 1, r, at + 1);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", V + i, W + i);
    st[i] = 0;
    idx[i] = i;
  }
  scanf("%d", &nq);
  for (int i = 0; i < n; ++i) {
    ed[i] = n + nq - 1;
  }
  int tot = n;
  for (int i = 0; i < nq; ++i) {
    scanf("%d", op + i + n);
    op[i + n]--;
    if (op[i + n] == 0) {
      scanf("%d%d", V + i + n, W + i + n);
      st[i + n] = i + n;
      ed[i + n] = n + nq - 1;
      idx[tot++] = n + i;
    } else if (op[i + n] == 1) {
      int x;
      scanf("%d", &x);
      x--;
      ed[idx[x]] = i + n;
    } else {
    }
  }
  for (int i = 0; i < n + nq; ++i) {
    if (op[i] == 0) {
      insert(st[i], ed[i], i, 0, n + nq - 1);
    }
  }
  solve(0, n + nq - 1, 1);
  for (int i = n; i < n + nq; ++i) {
    if (op[i] == 2) printf("%d\n", answer[i]);
  }
}