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>
int MOD = int(1e9) + 7;
inline int add(int a, int b) {
  return (a + b >= MOD) ? (a + b - MOD) : (a + b);
}
inline void inc(int &a, int b) { a = add(a, b); }
inline int sub(int a, int b) { return (a - b < 0) ? (a - b + MOD) : (a - b); }
inline void dec(int &a, int b) { a = sub(a, b); }
inline int mul(int a, int b) { return (a * 1ll * b) % MOD; }
using namespace std;
const int N = 15000, P = 10000019;
int V[N], W[N], from[N], to[N];
vector<int> ans;
void solve(int left, int right, bitset<N> used, vector<int> dp) {
  for (int i = 0; i < N; i++)
    if (!used[i] && from[i] <= left && to[i] >= right) {
      used[i] = 1;
      for (int j = ((int)(dp).size()) - 1; j >= W[i]; j--)
        dp[j] = max(dp[j], dp[j - W[i]] + V[i]);
    }
  if (left == right) {
    int x = 0;
    for (int i = ((int)(dp).size()) - 1; i > 0; i--) x = add(mul(x, P), dp[i]);
    ans.push_back(x);
  } else {
    int middle = (left + right) / 2;
    solve(left, middle, used, dp);
    solve(middle + 1, right, used, dp);
  }
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d %d", &V[i], &W[i]);
  fill(from, from + N, 0);
  fill(to, to + N, INT_MAX);
  int q, queries = 0;
  scanf("%d", &q);
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d %d", &V[n], &W[n]);
      from[n++] = queries;
    } else if (t == 2) {
      int x;
      scanf("%d", &x);
      to[--x] = queries - 1;
    } else
      queries++;
  }
  solve(0, queries - 1, bitset<N>(), vector<int>(k + 1, 0));
  for (int i : ans) printf("%d\n", i);
  return 0;
}