Problem: — Willem...

— What's the matter?

— It seems that there's something wrong with Seniorious...

— I'll have a look...

<image>

Seniorious is made by linking special talismans in particular order.

After over 500 years, the carillon is now in bad condition, so Willem decides to examine it thoroughly.

Seniorious has n pieces of talisman. Willem puts them in a line, the i-th of which is an integer ai.

In order to maintain it, Willem needs to perform m operations.

There are four types of operations:

  * 1 l r x: For each i such that l ≤ i ≤ r, assign ai + x to ai.
  * 2 l r x: For each i such that l ≤ i ≤ r, assign x to ai.
  * 3 l r x: Print the x-th smallest number in the index range [l, r], i.e. the element at the x-th position if all the elements ai such that l ≤ i ≤ r are taken and sorted into an array of non-decreasing integers. It's guaranteed that 1 ≤ x ≤ r - l + 1.
  * 4 l r x y: Print the sum of the x-th power of ai such that l ≤ i ≤ r, modulo y, i.e. <image>.

Input

The only line contains four integers n, m, seed, vmax (1 ≤ n, m ≤ 105, 0 ≤ seed < 109 + 7, 1 ≤ vmax ≤ 109).

The initial values and operations are generated using following pseudo code:
    
    
      
    def rnd():  
      
        ret = seed  
        seed = (seed * 7 + 13) mod 1000000007  
        return ret  
      
    for i = 1 to n:  
      
        a[i] = (rnd() mod vmax) + 1  
      
    for i = 1 to m:  
      
        op = (rnd() mod 4) + 1  
        l = (rnd() mod n) + 1  
        r = (rnd() mod n) + 1  
      
        if (l > r):   
             swap(l, r)  
      
        if (op == 3):  
            x = (rnd() mod (r - l + 1)) + 1  
        else:  
            x = (rnd() mod vmax) + 1  
      
        if (op == 4):  
            y = (rnd() mod vmax) + 1  
      
    

Here op is the type of the operation mentioned in the legend.

Output

For each operation of types 3 or 4, output a line containing the answer.

Examples

Input

10 10 7 9


Output

2
1
0
3


Input

10 10 9 9


Output

1
1
3
3

Note

In the first example, the initial array is {8, 9, 7, 2, 3, 1, 5, 6, 4, 8}.

The operations are:

  * 2 6 7 9
  * 1 3 10 8
  * 4 4 6 2 4
  * 1 4 5 8
  * 2 1 7 1
  * 4 7 9 4 4
  * 1 2 7 9
  * 4 5 8 1 1
  * 2 5 7 5
  * 4 3 10 8 5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
struct Domain {
  LL l, r;
  LL v;
  Domain(const LL &_l, const LL &_r, const LL &_v) : l(_l), r(_r), v(_v) {}
  bool operator<(const Domain &o) const { return l < o.l; }
};
ostream &operator<<(ostream &_cout, const Domain &d) {
  _cout << '[' << d.l << ' ' << d.v << ' ' << d.r << ']';
  return _cout;
}
LL N, M, seed, v;
set<Domain> domains;
vector<pair<LL, LL> > vec;
LL rnd() {
  LL ret = seed;
  seed = (seed * 7LL + 13) % 1000000007LL;
  return ret;
}
LL modPower(LL num, LL p, const LL &mod) {
  num %= mod;
  LL res = 1;
  do {
    if (p & 1) (res *= num) %= mod;
    (num *= num) %= mod;
  } while (p >>= 1);
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N >> M >> seed >> v;
  for (LL i = 1; i <= N; i++) {
    domains.emplace(i, i, (rnd() % v) + 1);
  }
  for (LL i = 1; i <= M; i++) {
    LL op, l, r, x, y;
    op = (rnd() % 4) + 1;
    l = (rnd() % N) + 1;
    r = (rnd() % N) + 1;
    if (l > r) swap(l, r);
    if (op == 3)
      x = (rnd() % (r - l + 1)) + 1;
    else
      x = (rnd() % v) + 1;
    if (op == 4) y = (rnd() % v) + 1;
    auto it = --domains.upper_bound(Domain(l, 0, 0));
    if (it->l < l) {
      LL tl = it->l, tr = it->r, tmp = it->v;
      domains.erase(it);
      domains.emplace(tl, l - 1, tmp);
      domains.emplace(l, tr, tmp);
    }
    it = --domains.upper_bound(Domain(r, 0, 0));
    if (it->r > r) {
      LL tl = it->l, tr = it->r, tmp = it->v;
      domains.erase(it);
      domains.emplace(tl, r, tmp);
      domains.emplace(r + 1, tr, tmp);
    }
    if (op == 1) {
      it = --domains.upper_bound(Domain(l, 0, 0));
      for (; it != domains.end() && it->r <= r; it++) {
        LL tl = it->l, tr = it->r;
        LL tmp = it->v;
        domains.erase(it);
        it = domains.emplace(tl, tr, tmp + x).first;
      }
    } else if (op == 2) {
      it = domains.lower_bound(Domain(l, 0, 0));
      for (; it != domains.end() && it->r <= r;) {
        it = domains.erase(it);
      }
      domains.emplace(l, r, x);
    } else if (op == 3) {
      vec.clear();
      it = --domains.upper_bound(Domain(l, 0, 0));
      for (; it != domains.end() && it->r <= r; it++) {
        vec.emplace_back(it->v, it->r - it->l + 1);
      }
      sort(vec.begin(), vec.end());
      LL ans;
      for (auto &item : vec) {
        if (x <= item.second) {
          ans = item.first;
          break;
        }
        x -= item.second;
      }
      cout << ans << '\n';
    } else if (op == 4) {
      auto it = --domains.upper_bound(Domain(l, 0, 0));
      LL ans = 0;
      for (; it != domains.end() && it->l <= r; it++) {
        ans +=
            ((modPower(it->v, x, y) * ((min(r, it->r) - it->l + 1)) % y)) % y;
        ans %= y;
      }
      cout << ans << '\n';
    }
  }
  return 0;
}