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 = 15500;
int timer = 1;
int add[N], del[N];
int v[N], w[N];
int sz = 0;
void readAdd() {
  scanf("%d%d", &v[sz], &w[sz]);
  add[sz] = timer;
  sz++;
}
void readDel() {
  int x;
  scanf("%d", &x);
  x--;
  del[x] = timer;
}
int k;
const int M = 1005;
const int LOG = 20;
int s[LOG][M];
const int p = (int)1e7 + 19;
const int q = (int)1e9 + 7;
void solve(int d, int l, int r, vector<int> &segs) {
  if (l == r) {
    if (l == timer) return;
    long long ans = 0;
    int mx = 0;
    int pp = 1;
    for (int i = 1; i <= k; i++) {
      mx = max(mx, s[d][i]);
      ans = (ans + mx * 1LL * pp) % q;
      pp = (pp * 1LL * p) % q;
    }
    printf("%lld\n", ans);
    return;
  }
  int m = (l + r) / 2;
  vector<int> segs1;
  for (int i = 0; i <= k; i++) s[d + 1][i] = s[d][i];
  for (int i : segs) {
    if (add[i] > m) continue;
    if (del[i] > m && add[i] <= l) {
      for (int j = k - w[i]; j >= 0; j--) {
        s[d + 1][j + w[i]] = max(s[d + 1][j + w[i]], s[d + 1][j] + v[i]);
      }
      continue;
    }
    segs1.push_back(i);
  }
  solve(d + 1, l, m, segs1);
  vector<int> segs2;
  for (int i = 0; i <= k; i++) s[d + 1][i] = s[d][i];
  for (int i : segs) {
    if (del[i] <= m + 1) continue;
    if (add[i] <= m + 1 && del[i] > r) {
      for (int j = k - w[i]; j >= 0; j--) {
        s[d + 1][j + w[i]] = max(s[d + 1][j + w[i]], s[d + 1][j] + v[i]);
      }
      continue;
    }
    segs2.push_back(i);
  }
  solve(d + 1, m + 1, r, segs2);
}
int main(int, char **) {
  int n;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    readAdd();
  }
  int qs;
  scanf("%d", &qs);
  for (int i = 0; i < qs; i++) {
    int x;
    scanf("%d", &x);
    if (x == 1) {
      readAdd();
    } else if (x == 2)
      readDel();
    else {
      timer++;
    }
  }
  for (int i = 0; i < sz; i++)
    if (del[i] == 0) del[i] = timer;
  vector<int> segs;
  for (int i = 0; i < sz; i++) segs.push_back(i);
  solve(0, 1, timer, segs);
  return 0;
}
