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 = 35000 + 5;
const int mod = 1e9 + 7;
const int P = 1e7 + 19;
const int inf = 1e9 + 7;
int n, i, s, x, y, k, q, w, p;
int dp[1005], Q[N];
vector<pair<int, int> > v[4 * N];
vector<int> aa;
pair<pair<int, int>, pair<int, int> > A[N];
void add(int root, int bas, int son, int x, int y, int v, int v2) {
  if (bas > y or son < x) return;
  if (x <= bas and son <= y) {
    ::v[root].push_back(make_pair(v, v2));
    return;
  }
  add((root + root), bas, ((bas + son) / 2), x, y, v, v2);
  add((root + root + 1), ((bas + son) / 2) + 1, son, x, y, v, v2);
}
void solve(int root, int bas, int son) {
  int dp2[1005] = {0}, i, j;
  memcpy(dp2, dp, sizeof dp);
  for (i = 0; i < v[root].size(); i++)
    for (j = k - v[root][i].second; j >= 0; j--)
      dp[j + v[root][i].second] =
          max(dp[j + v[root][i].second], dp[j] + v[root][i].first);
  if (bas == son) {
    if (Q[bas]) {
      int ans = 0;
      p = 1;
      for (i = 1; i <= k; i++) {
        ans = (ans + (long long)dp[i] * p % mod) % mod;
        p = (long long)p * P % mod;
      }
      aa.push_back(ans);
    }
    memcpy(dp, dp2, sizeof dp);
    return;
  }
  solve((root + root), bas, ((bas + son) / 2));
  solve((root + root + 1), ((bas + son) / 2) + 1, son);
  memcpy(dp, dp2, sizeof dp);
}
int main() {
  scanf("%d %d", &n, &k);
  for (i = 1; i <= n; i++) {
    scanf("%d %d", &A[i].first.first, &A[i].first.second);
    A[i].second.second = N;
  }
  scanf("%d", &q);
  s = n;
  for (i = 1; i <= q; i++) {
    scanf("%d", &w);
    if (w == 1) {
      scanf("%d %d", &x, &y);
      A[++s].first.first = x;
      A[s].first.second = y;
      A[s].second.first = i;
      A[s].second.second = N;
    } else if (w == 2) {
      scanf("%d", &x);
      A[x].second.second = i - 1;
    } else
      Q[i] = 1;
  }
  for (i = 1; i <= s; i++)
    add(1, 0, N, A[i].second.first, A[i].second.second, A[i].first.first,
        A[i].first.second);
  solve(1, 0, N);
  for (i = 0; i < aa.size(); i++) printf("%d\n", aa[i]);
}
