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 long long M = 1e9 + 7, P = 1e7 + 19;
const int N = 200005;
int nt = 1, k, n, Q, x, y, tot, op;
long long pm(long long x, long long p) {
  long long res = 1;
  while (p) {
    if (p & 1) res = res * x % M;
    p >>= 1;
    x = x * x % M;
  }
  return res;
}
struct node {
  int l, r, v, w;
} q[N << 2];
int qc[N];
vector<int> a;
long long f[50][2000];
void cdq(int l, int r, int d, vector<int> a) {
  if (d > 1) {
    for (int i = 1; i <= k; i++) f[d][i] = f[d - 1][i];
  }
  for (int i = 0; i < (int)a.size(); ++i)
    if (q[a[i]].l <= l && r <= q[a[i]].r) {
      for (int j = k; j >= q[a[i]].w; j--) {
        f[d][j] = max(f[d][j], f[d][j - q[a[i]].w] + q[a[i]].v);
      }
    }
  if (l == r) {
    long long res = 0;
    for (int i = k; i >= 1; i--) {
      res = (res * P + f[d][i]) % M;
    }
    for (int i = 1; i <= qc[l]; i++) {
      printf("%I64d\n", res);
    }
    return;
  }
  int mid = l + r >> 1;
  vector<int> la, ra;
  for (int i = 0; i < (int)a.size(); ++i) {
    if (q[a[i]].l <= l && r <= q[a[i]].r) continue;
    if (q[a[i]].r <= mid) {
      la.push_back(a[i]);
    } else if (q[a[i]].l > mid) {
      ra.push_back(a[i]);
    } else {
      la.push_back(a[i]);
      ra.push_back(a[i]);
    }
  }
  cdq(l, mid, d + 1, la);
  cdq(mid + 1, r, d + 1, ra);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &x, &y);
    q[++tot] = (node){1, 0, x, y};
  }
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d", &op);
    if (op == 1) {
      ++tot;
      ++nt;
      scanf("%d%d", &x, &y);
      q[tot] = (node){nt, 0, x, y};
    } else if (op == 2) {
      scanf("%d", &x);
      q[x].r = nt;
      ++nt;
    } else {
      qc[nt]++;
    }
  }
  for (int i = 1; i <= tot; i++) {
    a.push_back(i);
    if (q[i].r == 0) q[i].r = nt;
  }
  cdq(1, nt, 1, a);
  return 0;
}
