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 N = 15001, LOG = 20, K = 1001;
const int p = 10000000 + 19, q = 1000000000 + 7;
int v[N];
int w[N];
int b[N];
int e[N];
int dp[N][K];
long long pw[K];
int n, k, d;
const int offset = 1 << 15;
vector<int> tree[offset << 1];
int ans[offset];
void insert(int k, int a, int b) {
  a += offset;
  b += offset;
  while (a < b) {
    if (a & 1) tree[a++].push_back(k);
    if ((b & 1) == 0) tree[b--].push_back(k);
    a >>= 1;
    b >>= 1;
  }
  if (a == b) tree[a].push_back(k);
}
void dfs(int nr, int a, int b) {
  for (auto x : tree[nr]) {
    copy(dp[d], dp[d] + k + 1, dp[d + 1]);
    for (int i = k; i >= w[x]; i--)
      dp[d + 1][i] = max(dp[d + 1][i], dp[d][i - w[x]] + v[x]);
    d++;
  }
  if (a + 1 == b) {
    long long res = 0;
    for (int i = 1; i <= k; i++) res = (res + dp[d][i] * pw[i - 1]) % q;
    ans[a] = res;
  } else {
    dfs(nr * 2, a, (a + b) / 2);
    dfs(nr * 2 + 1, (a + b) / 2, b);
  }
  d -= tree[nr].size();
}
int main() {
  scanf("%d %d", &n, &k);
  pw[0] = 1;
  for (int i = 1; i <= k; i++) pw[i] = pw[i - 1] * p % q;
  for (int i = 1; i <= n; i++) scanf("%d %d", v + i, w + i);
  int q;
  vector<int> th;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int type;
    scanf("%d", &type);
    if (type == 3)
      th.push_back(i);
    else if (type == 2) {
      int x;
      scanf("%d", &x);
      e[x] = i;
    } else {
      n++;
      scanf("%d %d", v + n, w + n);
      b[n] = i;
    }
  }
  for (int i = 1; i <= n; i++) insert(i, b[i], e[i] ? e[i] : offset - 1);
  dfs(1, 0, offset);
  for (auto t : th) printf("%d\n", ans[t]);
}