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;
mt19937 Rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
template <typename T>
inline void chkmax(T &x, T y) {
  if (x < y) x = y;
}
template <typename T>
inline void chkmin(T &x, T y) {
  if (x > y) x = y;
}
inline int read() {
  int x = 0;
  char c = getchar();
  while (c < 48) c = getchar();
  while (c > 47) x = x * 10 + (c ^ 48), c = getchar();
  return x;
}
const int maxn = 35010, base = 1e7 + 19, P = 1e9 + 7;
int n, m, q, tot, pw[1010];
int st[35000], ed[35000];
struct node {
  int val, cost;
} A[35000];
int dp[17][1010];
vector<int> vec[maxn << 2];
void add(int k, int l, int r, int ql, int qr, int x) {
  if (ql <= l && r <= qr) return vec[k].push_back(x), void();
  if (ql <= ((l + r) >> 1)) add(k << 1, l, ((l + r) >> 1), ql, qr, x);
  if (qr > ((l + r) >> 1)) add(k << 1 | 1, ((l + r) >> 1) + 1, r, ql, qr, x);
}
void divide(int k, int l, int r, int dep) {
  memcpy(dp[dep], dp[dep - 1], (m + 1) << 2);
  for (int pos : vec[k]) {
    int val = A[pos].val, cost = A[pos].cost;
    for (int i = (m), iend = (cost); i >= iend; --i)
      chkmax(dp[dep][i], dp[dep][i - cost] + val);
  }
  if (l == r) {
    int res = 0;
    for (int i = (1), iend = (m); i <= iend; ++i)
      res = (res + 1ll * dp[dep][i] * pw[i - 1]) % P;
    printf("%d\n", res);
    return;
  }
  divide(k << 1, l, ((l + r) >> 1), dep + 1),
      divide(k << 1 | 1, ((l + r) >> 1) + 1, r, dep + 1);
}
int main() {
  pw[0] = 1;
  for (int i = (1), iend = (1000); i <= iend; ++i)
    pw[i] = 1ll * base * pw[i - 1] % P;
  cin >> n >> m;
  for (int i = (1), iend = (n); i <= iend; ++i) {
    A[i].val = read(), A[i].cost = read(), st[i] = 1;
  }
  memset(ed, 0x3f, sizeof ed);
  tot = n;
  cin >> q;
  int now = 0;
  for (int i = (1), iend = (q); i <= iend; ++i) {
    int op = read();
    if (op == 1) {
      tot++, A[tot].val = read(), A[tot].cost = read(), st[tot] = now + 1;
    } else if (op == 2) {
      ed[read()] = now;
    } else {
      now++;
    }
  }
  for (int i = (1), iend = (tot); i <= iend; ++i)
    if (st[i] <= ed[i]) {
      add(1, 1, now, st[i], ed[i], i);
    }
  divide(1, 1, now, 1);
  return 0;
}
