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;
long long seed, vmax;
set<pair<long long, long long> > Sx;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
long long getv(long long x) {
  auto it = Sx.lower_bound(pair<long long, long long>(x + 1, -1));
  it--;
  return it->second;
}
long long bpow(long long a, long long b, long long m) {
  a %= m;
  long long rv = 1 % m;
  while (b) {
    if (b & 1) rv = (rv * a) % m;
    a = (a * a) % m;
    b /= 2;
  }
  return rv;
}
int main() {
  long long n, m, i;
  scanf("%lld %lld %lld %lld", &n, &m, &seed, &vmax);
  for (i = 1; i <= n; ++i) {
    long long t = rnd() % vmax + 1;
    Sx.emplace(i, t);
  }
  vector<pair<long long, long long> > Vu, Vu2;
  for (int i = 1; i <= m; ++i) {
    long long op = rnd() % 4 + 1;
    long long L = rnd() % n + 1;
    long long R = rnd() % n + 1;
    if (L > R) swap(L, R);
    long long X, Y;
    if (op == 3)
      X = rnd() % (R - L + 1) + 1;
    else
      X = rnd() % vmax + 1;
    if (op == 4) Y = rnd() % vmax + 1;
    Vu.clear();
    long long v1 = getv(L), v2 = getv(R + 1);
    Sx.emplace(L, v1);
    if (R != n) Sx.emplace(R + 1, v2);
    while (1) {
      auto it = Sx.lower_bound(pair<long long, long long>(L, 0));
      if (it == Sx.end() or it->first > R) break;
      Vu.push_back(*it);
      Sx.erase(it);
    }
    if (op == 1) {
      for (auto it : Vu) Sx.emplace(it.first, it.second + X);
    } else if (op == 2) {
      Sx.emplace(L, X);
    } else if (op == 3) {
      for (auto it : Vu) Sx.insert(it);
      Vu.emplace_back(R + 1, 0);
      Vu2.clear();
      for (int i = 0; i + 1 < Vu.size(); ++i)
        Vu2.emplace_back(Vu[i].second, Vu[i + 1].first - Vu[i].first);
      sort(Vu2.begin(), Vu2.end());
      for (auto it : Vu2) {
        if (it.second >= X) {
          printf("%lld\n", it.first);
          break;
        } else
          X -= it.second;
      }
    } else if (op == 4) {
      for (auto it : Vu) Sx.insert(it);
      Vu.emplace_back(R + 1, 0);
      Vu2.clear();
      for (int i = 0; i + 1 < Vu.size(); i++)
        Vu2.emplace_back(Vu[i].second, Vu[i + 1].first - Vu[i].first);
      long long ans = 0;
      for (auto it : Vu2) (ans += bpow(it.first, X, Y) * it.second) %= Y;
      printf("%lld\n", ans);
    }
  }
}