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>
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long double eps = 1e-7;
const int inf = 1000000010;
const long long INF = 10000000000000010LL;
const int P = 1e7 + 19, Q = 1e9 + 7;
const int MAXN = 30010, K = 1001;
int n, q, k, u, v, x, y, t, a, b, ans;
int T[MAXN];
pair<int, int> A[MAXN];
int B[MAXN];
bool dead[MAXN];
int dp[K];
vector<pair<int, int> > seg[MAXN << 2];
void Add(int id, int tl, int tr, int l, int r, pair<int, int> p) {
  if (r <= tl || tr <= l) return;
  if (l <= tl && tr <= r) {
    seg[id].push_back(p);
    return;
  }
  int mid = (tl + tr) >> 1;
  Add(id << 1, tl, mid, l, r, p);
  Add(id << 1 | 1, mid, tr, l, r, p);
}
void add(int w, int v) {
  for (int i = k; i >= v; i--) dp[i] = max(dp[i], dp[i - v] + w);
}
void print() {
  long long res = 0;
  for (int m = k; m; m--) res = (res * P + dp[m]) % Q;
  cout << res << '\n';
}
void dfs(int id, int tl, int tr) {
  int dpp[K];
  memcpy(dpp, dp, sizeof(dp));
  for (pair<int, int> p : seg[id]) add(p.first, p.second);
  if (tr - tl == 1) {
    if (T[tl] == 3) print();
  } else {
    int mid = (tl + tr) >> 1;
    dfs(id << 1, tl, mid);
    dfs(id << 1 | 1, mid, tr);
  }
  memcpy(dp, dpp, sizeof(dp));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> A[i].first >> A[i].second;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> T[i];
    if (T[i] == 1) {
      n++;
      cin >> A[n].first >> A[n].second;
      B[n] = i;
      continue;
    }
    if (T[i] == 2) {
      cin >> x;
      Add(1, 0, q + 1, B[x], i, A[x]);
      dead[x] = 1;
    }
  }
  for (int i = 1; i <= n; i++)
    if (!dead[i]) Add(1, 0, q + 1, B[i], q + 1, A[i]);
  dfs(1, 0, q + 1);
  return 0;
}