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;
template <typename T>
inline void checkMin(T &a, T b) {
  if (b < a) a = b;
}
template <typename T>
inline void checkMax(T &a, T b) {
  if (a < b) a = b;
}
const int P = 1e7 + 19;
const int Q = 1e9 + 7;
const int MAX_N = 5005 + 30000;
int n, k, w[MAX_N], v[MAX_N];
long long dp[32][1001];
int beg[MAX_N], qc[MAX_N];
set<pair<int, int>, greater<pair<int, int>>> es[MAX_N];
void dfs(int depth, int l, int r) {
  copy(dp[depth - 1], dp[depth], dp[depth]);
  for (auto iter = es[l].begin(); iter != es[l].end();) {
    if (iter->first < r) break;
    int i = iter->second;
    if (iter->first > r) {
      es[r + 1].emplace(iter->first, i);
    }
    iter = es[l].erase(iter);
    for (int j = k; j >= w[i]; --j) {
      checkMax(dp[depth][j], dp[depth][j - w[i]] + v[i]);
    }
  }
  if (l == r) {
    if (!qc[l]) return;
    int ans = 0;
    for (int i = k; i >= 1; --i) {
      ans = ((long long)ans * P + dp[depth][i]) % Q;
    }
    for (int i = 0; i < int(qc[l]); ++i) {
      printf("%d\n", ans);
    }
  } else {
    int m = (l + r) >> 1;
    dfs(depth + 1, l, m);
    dfs(depth + 1, m + 1, r);
  }
}
int main() {
  while (~scanf("%d%d", &n, &k)) {
    fill(qc, qc + MAX_N, 0);
    fill(es, es + MAX_N, set<pair<int, int>, greater<pair<int, int>>>());
    int ts = 1;
    for (int i = 1; i <= n; ++i) {
      scanf("%d%d", &v[i], &w[i]);
      beg[i] = ts;
    }
    int q, type, x;
    scanf("%d", &q);
    for (int i = 0; i < int(q); ++i) {
      scanf("%d", &type);
      if (type == 1) {
        ++ts;
        ++n;
        scanf("%d%d", &v[n], &w[n]);
        beg[n] = ts;
      } else if (type == 2) {
        scanf("%d", &x);
        es[beg[x]].emplace(ts++, x);
        beg[x] = -1;
      } else {
        ++qc[ts];
      }
    }
    for (int i = 1; i <= n; ++i) {
      if (beg[i] == -1) continue;
      es[beg[i]].emplace(ts, i);
    }
    dfs(1, 1, ts);
  }
  return 0;
}