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 maxn = 60000 + 10;
const int maxk = 2005;
const long long seed = (int)1e7 + 19;
const long long mod = (int)1e9 + 7;
struct info {
  int t, p;
  long long v;
};
stack<info> chg;
long long dp[maxk], ans[maxn];
int n, k, q;
pair<int, int> ask[maxn];
int v[maxn], w[maxn], vis[maxn];
void modify(int tm, int pos, long long val) {
  chg.push((info){tm, pos, dp[pos]});
  dp[pos] = val;
}
void goback(int tm) {
  while (!chg.empty() && chg.top().t > tm) {
    dp[chg.top().p] = chg.top().v;
    chg.pop();
  }
}
void add(int tm, int v, int w) {
  for (int i = k; i >= w; i--) {
    if (dp[i - w] + v > dp[i]) {
      modify(tm, i, dp[i - w] + v);
    }
  }
}
void prepl(int tm, int l, int r) {
  for (int i = l; i <= r; i++) {
    if (ask[i].first == 1) vis[ask[i].second] = tm;
    if (ask[i].first == 2 && i >= ((l + r) >> 1) + 1) {
      if (vis[ask[i].second] != tm) add(tm, v[ask[i].second], w[ask[i].second]);
    }
  }
}
void prepr(int tm, int l, int r) {
  for (int i = r; i >= l; i--) {
    if (ask[i].first == 2) vis[ask[i].second] = tm;
    if (ask[i].first == 1 && i <= ((l + r) >> 1)) {
      if (vis[ask[i].second] != tm) add(tm, v[ask[i].second], w[ask[i].second]);
    }
  }
}
void solve(int cur, int l, int r) {
  if (l == r) {
    if (ask[l].first == 3) {
      ans[l] = 0;
      for (int i = k; i > 0; i--) {
        ans[l] = (1ll * ans[l] * seed % mod + dp[i] % mod) % mod;
      }
    }
    return;
  }
  prepl((cur << 1), l, r);
  solve((cur << 1), l, ((l + r) >> 1));
  goback(cur);
  prepr((cur << 1 | 1), l, r);
  solve((cur << 1 | 1), ((l + r) >> 1) + 1, r);
  goback(cur);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &v[i], &w[i]);
    ask[i].first = 1;
    ask[i].second = i;
  }
  int tmpn = n;
  scanf("%d", &q);
  for (int i = tmpn + 1; i <= q + tmpn; i++) {
    scanf("%d", &ask[i].first);
    if (ask[i].first == 1) {
      ++n;
      scanf("%d%d", &v[n], &w[n]);
      ask[i].second = n;
    }
    if (ask[i].first == 2) scanf("%d", &ask[i].second);
    if (ask[i].first == 3) ask[i].second = 0;
  }
  q += tmpn;
  while (!chg.empty()) chg.pop();
  memset(vis, 0, sizeof vis);
  memset(dp, 0, sizeof dp);
  solve(1, 1, q);
  for (int i = tmpn + 1; i <= q; i++)
    if (ask[i].first == 3) printf("%I64d\n", ans[i]);
  return 0;
}