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 = 5e3 + 3;
const int I = 1e4 + 4;
const int Q = 3e4 + 4;
const int K = 1e3 + 3;
const int logQ = 16;
const int mod = 1e9 + 7;
const int p = 1e7 + 19;
int n, k;
int v[N + I];
int w[N + I];
int idx;
int q;
int type[Q];
int val[Q];
int tin[N + I];
int tout[N + I];
int dp[logQ * K];
void print(int l, int r) {
  int cur = 1;
  int res = 0;
  for (int i = l; i <= r; ++i) {
    res = (res + 1LL * cur * dp[i]) % mod;
    cur = (1LL * cur * p) % mod;
  }
  printf("%d\n", res);
}
void solve(int l, int r, int pl, int pr, int last) {
  bool ok = 0;
  for (int i = l; i <= r; ++i) {
    if (type[i] == 3) {
      ok = 1;
      break;
    }
  }
  if (!ok) {
    return;
  }
  for (int i = 1; i <= k; ++i) {
    dp[last + i] = dp[last + i - k];
  }
  int mid = pl + pr >> 1;
  if (l == pl) {
    for (int i = mid + 1; i <= pr; ++i) {
      if (type[i] == 2 && tin[val[i]] < l) {
        int who = val[i];
        for (int j = k; j >= w[who]; --j) {
          dp[j + last] =
              max(dp[j + last],
                  ((j > w[who]) ? dp[j + last - w[who]] : 0) + v[who]);
        }
      }
    }
  } else {
    for (int i = pl; i <= mid; ++i) {
      if (type[i] == 1 && tout[val[i]] > r) {
        int who = val[i];
        for (int j = k; j >= w[who]; --j) {
          dp[j + last] =
              max(dp[j + last],
                  ((j > w[who]) ? dp[j + last - w[who]] : 0) + v[who]);
        }
      }
    }
  }
  if (l < r) {
    int mid = l + r >> 1;
    solve(l, mid, l, r, last + k);
    solve(mid + 1, r, l, r, last + k);
  } else {
    print(last + 1, last + k);
  }
}
int main() {
  scanf("%d %d", &n, &k);
  idx = 0;
  for (int i = 1; i <= n; ++i) {
    ++idx;
    scanf("%d %d", v + idx, w + idx);
    tin[idx] = 0;
    tout[idx] = Q;
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    scanf("%d", type + i);
    if (type[i] == 1) {
      ++idx;
      scanf("%d %d", v + idx, w + idx);
      val[i] = idx;
      tin[idx] = i;
      tout[idx] = Q;
    } else if (type[i] == 2) {
      scanf("%d", val + i);
      tout[val[i]] = i;
    }
  }
  for (int i = 1; i <= idx; ++i) {
    if (tout[i] == Q && tin[i] == 0) {
      for (int j = k; j >= w[i]; --j) {
        dp[j] = max(dp[j], dp[j - w[i]] + v[i]);
      }
    }
  }
  if (q == 1) {
    print(1, k);
  } else {
    solve(1, 1 + q >> 1, 1, q, k);
    solve((1 + q >> 1) + 1, q, 1, q, k);
  }
}
