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>
#pragma GCC optimize("Ofast")
#pragma GCC target( \
    "avx,avx2,fma,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long MOD = 1e9 + 7;
long long k;
vector<long long> v, w, num;
vector<pair<long long, pair<long long, long long> > > zp;
vector<vector<long long> > dp;
long long st(long long t, long long id, long long mod) {
  long long res = 1, a = t, n = id;
  while (n) {
    if (n & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    n >>= 1;
  }
  return res % mod;
}
void add(long long id) {
  vector<long long> mem(k + 1);
  dp.push_back(mem);
  long long lst = dp.size() - 1;
  for (int i = 0; i < k + 1; i++) {
    dp[lst][i] = dp[lst - 1][i];
    if (i >= w[id])
      dp[lst][i] = max(dp[lst - 1][i], dp[lst - 1][i - w[id]] + v[id]);
  }
}
void del() { dp.pop_back(); }
long long ans() {
  long long lst = dp.size() - 1;
  long long anss = 0;
  for (int i = 1; i < k + 1; i++) {
    anss = ((10000019 * (long long)anss) + (long long)dp[lst][k - i + 1]) % MOD;
    if (anss > MOD) anss -= MOD;
  }
  return anss;
}
void get(long long l, long long r) {
  if (r == l + 1) {
    if (zp[l].first == 3) {
      zp[l].second.first = ans();
    }
    return;
  }
  long long mid = (l + r) / 2;
  long long kol = 0;
  for (int i = mid; i < r; i++) {
    if (zp[i].first == 2) {
      if (num[zp[i].second.second] < l) {
        add(zp[i].second.second);
        kol++;
      }
    }
  }
  get(l, mid);
  while (kol--) del();
  kol = 0;
  for (int i = l; i < mid; i++) {
    if (zp[i].first == 1) {
      if (zp[i].second.second >= r) {
        add(zp[i].second.first);
        kol++;
      }
    }
  }
  get(mid, r);
  while (kol--) del();
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0),
      cout << fixed << setprecision(20);
  long long n;
  cin >> n >> k;
  v.resize(n), w.resize(n), num.resize(n);
  vector<long long> mem(k + 1);
  dp.push_back(mem);
  for (int i = 0; i < n; i++) {
    cin >> v[i] >> w[i];
    zp.push_back({1, {i, MOD}});
    num[i] = i;
  }
  long long q;
  cin >> q;
  while (q--) {
    long long sz = zp.size();
    zp.push_back({0, {num.size(), MOD}});
    cin >> zp[zp.size() - 1].first;
    if (zp.back().first == 1) {
      long long vv, ww;
      cin >> vv >> ww;
      v.push_back(vv);
      w.push_back(ww);
      num.push_back(sz);
    }
    if (zp.back().first == 2) {
      long long x;
      cin >> x;
      x--;
      zp[num[x]].second.second = sz;
      zp[sz].second.second = x;
    }
  }
  get(0, zp.size());
  for (auto i : zp) {
    if (i.first == 3) {
      cout << i.second.first << endl;
    }
  }
}
