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 long double pi = 3.1415926535897932384626433832795l;
template <typename T>
inline auto sqr(T x) -> decltype(x * x) {
  return x * x;
}
template <typename T1, typename T2>
inline bool umx(T1& a, T2 b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
inline bool umn(T1& a, T2 b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
struct Input {
  int n, k, q;
  vector<int> v, w;
  vector<int> tq, vq, wq;
  bool read() {
    if (scanf("%d%d", &n, &k) != 2) return 0;
    v.resize(n), w.resize(n);
    for (int i = int(0); i < int(n); ++i) {
      scanf("%d%d", &v[i], &w[i]);
    }
    scanf("%d", &q);
    tq.resize(q), vq = wq = tq;
    for (int i = int(0); i < int(q); ++i) {
      scanf("%d", &tq[i]);
      --tq[i];
      if (tq[i] == 0) {
        scanf("%d%d", &vq[i], &wq[i]);
      } else if (tq[i] == 1) {
        scanf("%d", &vq[i]);
        --vq[i];
      }
    }
    return 1;
  }
  void init(const Input& input) { *this = input; }
};
struct Data : Input {
  vector<int> ans;
  void write() {
    for (int x : ans) {
      printf("%d\n", x);
    }
  }
};
struct Solution : Data {
  static const int BASE = 10000019;
  static const int MOD = 1000000007;
  static const int MAXIT = 20;
  static const int MAXK = 1001;
  int dp[MAXIT][MAXK];
  vector<int> lpos, rpos;
  void rec(int l, int r, int it) {
    if (l == r - 1) {
      if (tq[l] == 2) {
        int res = 0;
        for (int j = int(k + 1) - 1; j >= int(1); --j) {
          res = (1ll * res * BASE + dp[it][j]) % MOD;
        }
        ans.push_back(res);
      }
    } else {
      int mid = (l + r) / 2;
      memcpy(dp[it + 1], dp[it], (k + 1) * sizeof(dp[it + 1][0]));
      for (int i = int(mid); i < int(r); ++i) {
        if (tq[i] == 1 && lpos[i] < l) {
          int id = vq[i];
          int tv = v[id], tw = w[id];
          for (int t = int(k + 1) - 1; t >= int(tw); --t) {
            umx(dp[it + 1][t], dp[it + 1][t - tw] + tv);
          }
        }
      }
      rec(l, mid, it + 1);
      memcpy(dp[it + 1], dp[it], (k + 1) * sizeof(dp[it + 1][0]));
      for (int i = int(l); i < int(mid); ++i) {
        if (tq[i] == 0 && rpos[i] >= r) {
          int id = vq[i];
          int tv = v[id], tw = w[id];
          for (int t = int(k + 1) - 1; t >= int(tw); --t) {
            umx(dp[it + 1][t], dp[it + 1][t - tw] + tv);
          }
        }
      }
      rec(mid, r, it + 1);
    }
  }
  void solve() {
    for (int i = int(0); i < int(q); ++i) {
      if (tq[i] == 0) {
        v.push_back(vq[i]);
        w.push_back(wq[i]);
        vq[i] = ((int)(v).size()) - 1;
      }
    }
    memset(dp, 0, sizeof(dp));
    {
      vector<int> ttt(n, 0), ttv(n);
      for (int i = int(0); i < int(n); ++i) {
        ttv[i] = i;
      }
      ttt.swap(tq);
      tq.insert(tq.end(), (ttt).begin(), (ttt).end());
      ttv.swap(vq);
      vq.insert(vq.end(), (ttv).begin(), (ttv).end());
      n = ((int)(v).size());
      q = ((int)(tq).size());
      vector<int> used(n, 0);
      for (int i = int(0); i < int(q); ++i) {
        if (tq[i] == 0) {
          used[vq[i]]++;
        } else if (tq[i] == 1) {
          used[vq[i]]--;
        }
      }
      for (int i = int(0); i < int(n); ++i) {
        if (used[i]) {
          tq.push_back(1);
          vq.push_back(i);
        }
      }
    }
    q = ((int)(tq).size());
    {
      vector<int> cnt(n, -1);
      lpos.resize(q, -1);
      for (int i = int(0); i < int(q); ++i) {
        if (tq[i] == 0) {
          cnt[vq[i]] = i;
        } else if (tq[i] == 1) {
          lpos[i] = cnt[vq[i]];
        }
      }
      cnt.assign(n, q + 1);
      rpos.resize(q, q + 1);
      for (int i = int(q) - 1; i >= int(0); --i) {
        if (tq[i] == 1) {
          cnt[vq[i]] = i;
        } else if (tq[i] == 0) {
          rpos[i] = cnt[vq[i]];
        }
      }
    }
    rec(0, q, 0);
  }
  void clear() { *this = Solution(); }
};
Solution sol;
int main() {
  cout.setf(ios::showpoint | ios::fixed);
  cout.precision(20);
  sol.read();
  sol.solve();
  sol.write();
  return 0;
}