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>
using namespace std;
const int MAXK = 1005, TOUR = 1 << 16, MOD = 1e9 + 7, PP = 1e7 + 19;
int n, k;
int dp[2][MAXK];
int inddp;
int poc[TOUR], kraj[TOUR], val[TOUR], masa[TOUR];
int ind = 0;
int jelq[TOUR];
vector<int> T[2 * TOUR];
int Add(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
int Mul(int a, int b) { return (int)(((long long)a * b) % MOD); }
int Eval() {
  int mnoz = 1, ret = 0;
  for (int i = 1; i <= k; i++) {
    ret = Add(ret, Mul(mnoz, dp[inddp][i]));
    mnoz = Mul(mnoz, PP);
  }
  return ret;
}
void Stavi(int pos, int lo, int hi, int begin, int end, int vv) {
  if (lo >= end || hi <= begin) return;
  if (lo >= begin && hi <= end) {
    T[pos].push_back(vv);
    return;
  }
  Stavi(2 * pos + 0, lo, (lo + hi) / 2, begin, end, vv);
  Stavi(2 * pos + 1, (lo + hi) / 2, hi, begin, end, vv);
}
void Dfs(int pos) {
  int tmp[2][MAXK];
  memcpy(tmp, dp, sizeof dp);
  int pocind = inddp;
  for (int i = 0; i < (int)T[pos].size(); i++, inddp ^= 1) {
    int a = T[pos][i];
    for (int j = 0; j <= k; j++) dp[1 ^ inddp][j] = dp[inddp][j];
    for (int j = 0; j <= k - masa[a]; j++)
      dp[1 ^ inddp][j + masa[a]] =
          max(dp[1 ^ inddp][j + masa[a]], dp[inddp][j] + val[a]);
  }
  if (pos < TOUR) {
    Dfs(2 * pos);
    Dfs(2 * pos + 1);
  } else if (jelq[pos - TOUR])
    printf("%d\n", Eval());
  inddp = pocind;
  memcpy(dp, tmp, sizeof dp);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &val[ind], &masa[ind]);
    ind++;
  }
  int brq;
  scanf("%d", &brq);
  for (int i = 1; i <= brq; i++) {
    int st;
    scanf("%d", &st);
    if (st == 1) {
      scanf("%d%d", &val[ind], &masa[ind]);
      poc[ind++] = i;
    } else if (st == 2) {
      int koji;
      scanf("%d", &koji);
      koji--;
      kraj[koji] = i;
    } else
      jelq[i] = 1;
  }
  for (int i = 0; i < ind; i++) {
    if (!kraj[i]) kraj[i] = brq + 1;
    Stavi(1, 0, TOUR, poc[i], kraj[i], i);
  }
  Dfs(1);
  return 0;
}