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 MAXN = 15010;
const int MAXK = 1010;
const int MAXQ = 30010;
const int modP = 1e7 + 19;
const int modQ = 1e9 + 7;
using namespace std;
int N, K, Q, cnt;
int W[MAXN], V[MAXN];
vector<int> tree[MAXQ * 4];
map<int, int> mp;
vector<int> queries;
long long pot[MAXK], ans[MAXQ];
long long dp[MAXN][MAXK];
int m(int l, int r) { return (l + r) >> 1; }
void upd(int pos, int l, int r, int beg, int en, int x) {
  if (l > en || r < beg) return;
  if (l >= beg && r <= en) return (void)(tree[pos].push_back(x));
  upd((pos << 1), l, m(l, r), beg, en, x);
  upd((pos << 1) | 1, m(l, r) + 1, r, beg, en, x);
}
void run(int pos, int l, int r, int last) {
  ("Run %d %d %d %d\n", pos, l, r, last);
  for (auto idx : tree[pos]) {
    ("With item %d %d %d\n", idx, V[idx], W[idx]);
    last++;
    int v = V[idx], w = W[idx];
    for (int i = 1; i <= K; i++) {
      dp[last][i] = dp[last - 1][i];
      if (i - w >= 0)
        dp[last][i] = max(dp[last - 1][i], dp[last - 1][i - w] + v);
    }
  }
  for (int i = 1; i <= K; i++) ("%lld ", dp[last][i]);
  ("\n\n");
  if (l == r) {
    long long mx = -1;
    for (int i = 1; i <= K; i++) {
      mx = max(mx, dp[last][i]);
      ans[l] = (ans[l] + (pot[i - 1] * mx) % modQ) % modQ;
    }
    return;
  }
  run(pos << 1, l, m(l, r), last);
  run((pos << 1) | 1, m(l, r) + 1, r, last);
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; i++) {
    scanf("%d%d", &V[i], &W[i]);
    mp.insert(make_pair(i, 1));
  }
  scanf("%d", &Q);
  cnt = 1 + N;
  for (int i = 1; i <= Q; i++) {
    int type, x;
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d%d", &V[cnt], &W[cnt]);
      mp.insert(make_pair(cnt, i));
      cnt++;
    }
    if (type == 2) {
      scanf("%d", &x);
      upd(1, 1, Q, mp[x], i, x);
      mp.erase(mp.find(x));
    } else if (type == 3)
      queries.push_back(i);
  }
  for (auto p : mp) upd(1, 1, Q, p.second, Q, p.first);
  pot[0] = 1LL;
  for (int i = 1; i <= K; i++) pot[i] = (pot[i - 1] * modP) % modQ;
  run(1, 1, Q, 0);
  for (int i : queries) printf("%lld\n", ans[i]);
}