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;
const int N = 30003, K = 1009, LG = 17, Mod = 1e9 + 7;
int n, q, k, m, V[N], W[N], L[N], R[N], Pw[N], dp[LG][K], res[N];
vector<int> Vec[N * 4];
void Add(int le, int ri, int i, int id = 1, int l = 0, int r = m) {
  if (ri <= l || r <= le) return;
  if (le <= l && r <= ri) return void(Vec[id].push_back(i));
  Add(le, ri, i, (id << 1), l, (l + r >> 1));
  Add(le, ri, i, ((id << 1) ^ 1), (l + r >> 1), r);
}
void DFS(int id = 1, int l = 0, int r = m, int h = 1) {
  for (int i = 0; i <= k; i++) dp[h][i] = dp[h - 1][i];
  for (int i : Vec[id])
    for (int j = k; j >= W[i]; j--)
      dp[h][j] = max(dp[h][j], dp[h][j - W[i]] + V[i]);
  if (r - l < 2) {
    int hsh = 0, Mx = dp[h][0];
    for (int i = 1; i <= k; i++)
      Mx = max(Mx, dp[h][i]), hsh = (hsh + Mx * 1LL * Pw[i - 1]) % Mod;
    res[l] = hsh;
    return;
  } else
    DFS((id << 1), l, (l + r >> 1), h + 1),
        DFS(((id << 1) ^ 1), (l + r >> 1), r, h + 1);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d%d", &V[i], &W[i]), L[i] = 0, R[i] = N;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int id, tp;
    scanf("%d", &tp);
    if (tp == 1)
      scanf("%d%d", &V[n], &W[n]), L[n] = m, R[n] = N, n++;
    else if (tp == 2)
      scanf("%d", &id), R[id - 1] = m;
    else
      m++;
  }
  Pw[0] = 1;
  for (int i = 1; i < N; i++) Pw[i] = Pw[i - 1] * (long long)(1e7 + 19) % Mod;
  memset(dp, -63, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) Add(L[i], min(R[i], m), i);
  DFS();
  for (int i = 0; i < m; i++) printf("%d\n", res[i]);
  return 0;
}
