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;
inline void Max(int &a, int b) {
  if (b > a) a = b;
}
const int N = (int)5e3 + 5, MAX_K = (int)1e3 + 5, Q = (int)3e4 + 5, T = N + Q,
          p = (int)1e7 + 19, q = (int)1e9 + 7;
int K, pow_p[MAX_K];
vector<pair<int, int> > vec[T << 2];
struct Item {
  int cost, val, L, R;
  Item() {}
  Item(int cost, int val, int L, int R) : cost(cost), val(val), L(L), R(R) {}
} item[N + Q];
void ins(int L, int R, int l, int r, int cost, int val, int k = 1) {
  if (L == l && R == r) {
    vec[k].push_back(pair<int, int>(cost, val));
    return;
  }
  int mid = L + R >> 1;
  if (r <= mid)
    ins(L, mid, l, r, cost, val, (k << 1));
  else if (l > mid)
    ins(mid + 1, R, l, r, cost, val, (k << 1 | 1));
  else {
    ins(L, mid, l, mid, cost, val, (k << 1));
    ins(mid + 1, R, mid + 1, r, cost, val, (k << 1 | 1));
  }
}
void rec(int L, int R, int dep, int k = 1) {
  static int dp[20][MAX_K];
  memcpy(dp[dep], dp[dep - 1], K + 1 << 2);
  for (pair<int, int> elm : vec[k]) {
    int cost = elm.first, val = elm.second;
    for (int i = (K + 1) - 1, _s = (cost); i >= _s; --i)
      Max(dp[dep][i], dp[dep][i - cost] + val);
  }
  if (L == R) {
    int ans = 0;
    for (int i = (1), _t = (K + 1); i < _t; ++i)
      ans = (ans + (long long)dp[dep][i] * pow_p[i - 1]) % q;
    printf("%d\n", ans);
    return;
  }
  int mid = L + R >> 1;
  rec(L, mid, dep + 1, (k << 1));
  rec(mid + 1, R, dep + 1, (k << 1 | 1));
}
int main() {
  int n, q, cur = 1;
  cin >> n >> K;
  pow_p[0] = 1;
  for (int i = (1), _t = (K + 1); i < _t; ++i)
    pow_p[i] = (long long)pow_p[i - 1] * p % (::q);
  for (int i = (1), _t = (n + 1); i < _t; ++i) {
    scanf("%d%d", &item[i].val, &item[i].cost);
    item[i].L = 1;
    item[i].R = -1;
  }
  cin >> q;
  for (int i = 1, tar, type, id = n + 1; i <= q; ++i) {
    scanf("%d", &type);
    switch (type) {
      case 1:
        scanf("%d%d", &item[id].val, &item[id].cost);
        item[id].L = cur;
        item[id].R = -1;
        ++id;
        break;
      case 2:
        scanf("%d", &tar);
        item[tar].R = cur - 1;
        break;
      case 3:
        ++cur;
        break;
    }
  }
  for (int i = (1), _t = (n + q + 1); i < _t; ++i)
    if (item[i].cost) {
      if (!~item[i].R) item[i].R = cur - 1;
      if (item[i].L <= item[i].R)
        ins(1, cur - 1, item[i].L, item[i].R, item[i].cost, item[i].val);
    }
  rec(1, cur - 1, 1);
  return 0;
}
