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;
inline int read() {
  char ch = getchar();
  int x = 0, op = 1;
  while (ch < '0' || '9' < ch) {
    if (ch == '-') op = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return op * x;
}
const int maxdep = 16, maxn = 30009;
const int base = 10000019, MOD = 1000000007;
int power[maxn], v[maxn], w[maxn], L[maxn], R[maxn], flag[maxn];
vector<pair<int, int> > T[maxn << 2];
long long dp[maxdep][1009];
int n, K, Q, m, t, a, b, ans;
void update(int o, int l, int r, int x, int y, int v, int w) {
  if (l == x && y == r) {
    T[o].push_back(make_pair(v, w));
    return;
  }
  if (x <= ((l + r) >> 1))
    update((o << 1), l, ((l + r) >> 1), x, min(y, ((l + r) >> 1)), v, w);
  if (((l + r) >> 1) + 1 <= y)
    update((o << 1 | 1), ((l + r) >> 1) + 1, r, max(((l + r) >> 1) + 1, x), y,
           v, w);
}
void solve(int o, int dep, int l, int r) {
  memset(dp[dep], 0, sizeof dp[dep]);
  if (dep)
    for (int i = (1); i <= (K); i++) dp[dep][i] = dp[dep - 1][i];
  if (!T[o].empty())
    for (int i = (0); i <= (T[o].size() - 1); i++)
      for (int j = (K); j >= (T[o][i].second); j--)
        dp[dep][j] =
            max(dp[dep][j], dp[dep][j - T[o][i].second] + T[o][i].first);
  if (l == r) {
    if (flag[l]) {
      ans = 0;
      for (int i = (1); i <= (K); i++)
        ans = (ans + 1LL * dp[dep][i] % MOD * power[i - 1]) % MOD;
      printf("%d\n", ans);
    }
    return;
  }
  solve((o << 1), dep + 1, l, ((l + r) >> 1));
  solve((o << 1 | 1), dep + 1, ((l + r) >> 1) + 1, r);
}
int main() {
  m = n = read();
  K = read();
  power[0] = 1;
  for (int i = (1); i <= (K); i++) power[i] = 1LL * power[i - 1] * base % MOD;
  for (int i = (1); i <= (n); i++) {
    v[i] = read();
    w[i] = read();
  }
  Q = read();
  for (int i = (1); i <= (n); i++) {
    L[i] = 1;
    R[i] = Q;
  }
  for (int i = (1); i <= (Q); i++) {
    t = read();
    if (t == 1) {
      v[++m] = read();
      w[m] = read();
      L[m] = i;
      R[m] = Q;
    } else if (t == 2)
      R[read()] = i;
    else
      flag[i] = 1;
  }
  for (int i = (1); i <= (m); i++) update(1, 1, Q, L[i], R[i], v[i], w[i]);
  solve(1, 0, 1, Q);
  return 0;
}
