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 int N = 1e5 + 500;
const int M = 1e6 + 500;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int LOG = 20;
const int ALP = 30;
const int BASE = 31337;
const double EPS = 1e-9;
const double PI = 3.1415926535;
const int P = 1e7 + 19;
const int OFF = (1 << 15);
inline int add(const int &A, const int &B) {
  if (A + B >= MOD) return A + B - MOD;
  return A + B;
}
inline int mul(const int &A, const int &B) { return (long long)A * B % MOD; }
inline int sub(const int &A, const int &B) {
  if (A - B < 0) return A - B + MOD;
  return A - B;
}
inline int pot(const int &A, int B) {
  int cur = A, ret = 1;
  for (; B; B >>= 1) {
    if (B & 1) ret = mul(ret, cur);
    cur = mul(cur, cur);
  }
  return ret;
}
inline int divv(const int &A, const int &B) { return mul(A, pot(B, MOD - 2)); }
vector<vector<long long> > dp;
vector<int> ans;
vector<pair<int, int> > v[2 * OFF];
int W, n, q, pre_pot[N], ispis[OFF], st[N], pos[N], un[N], sm[2 * OFF];
pair<int, int> novi[N];
void dodaj(pair<int, int> first) {
  int v = first.first, w = first.second;
  vector<long long> nw(W + 1, 0);
  int odg = 0;
  for (int i = 1; i <= W; i++) {
    if ((int)dp.size() > 0) nw[i] = dp.back()[i];
    if (i >= w)
      nw[i] = max(nw[i], ((int)dp.size() > 0 ? dp.back()[i - w] : 0) + v);
    odg = add(odg, mul(nw[i] % MOD, pre_pot[i - 1]));
  }
  dp.push_back(nw);
  ans.push_back(odg);
}
void makni(int k) {
  for (; k--;) dp.pop_back(), ans.pop_back();
}
void ubaci(int i, int a, int b, int lo, int hi, pair<int, int> first) {
  if (lo <= a && b <= hi) {
    v[i].push_back(first);
    return;
  }
  if (a > hi || b < lo) return;
  ubaci(2 * i, a, (a + b) / 2, lo, hi, first);
  ubaci(2 * i + 1, (a + b) / 2 + 1, b, lo, hi, first);
}
void obradi(int i) {
  if (!sm[i]) return;
  for (pair<int, int> tmp : v[i]) dodaj(tmp);
  if (i < OFF) {
    obradi(2 * i);
    obradi(2 * i + 1);
  } else if (ispis[i - OFF]) {
    if (ans.size() > 0)
      printf("%d\n", ans.back());
    else
      printf("0\n");
  }
  makni((int)v[i].size());
}
int main() {
  pre_pot[0] = 1;
  for (int i = 1; i < N; i++) pre_pot[i] = mul(pre_pot[i - 1], P);
  scanf("%d%d", &n, &W);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &novi[i].first, &novi[i].second);
    st[i] = 0;
    un[i] = 1;
  }
  scanf("%d", &q);
  int cur = n + 1;
  for (int i = 0; i < q; i++) {
    int ty;
    scanf("%d", &ty);
    if (ty == 3) {
      ispis[i] = 1;
      for (int ii = (OFF + i); ii; ii /= 2) sm[ii]++;
    } else if (ty == 2) {
      int x;
      scanf("%d", &x);
      un[x] = 0;
      ubaci(1, 0, OFF - 1, st[x], i, novi[x]);
    } else {
      st[cur] = i;
      un[cur] = 1;
      scanf("%d%d", &novi[cur].first, &novi[cur].second);
      cur++;
    }
  }
  for (int i = 0; i < N; i++)
    if (un[i]) ubaci(1, 0, OFF - 1, st[i], OFF - 1, novi[i]);
  obradi(1);
  return 0;
}
