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;
map<int, long long> mp;
int n, m, l, r, x, y, seed, maxVal, op;
vector<pair<long long, int>> v;
int rnd() {
  int s = seed;
  seed = (1LL * seed * 7 + 13) % 1000000007;
  return s;
}
long long pw(long long a, long long b) {
  return b ? b & 1 ? a * pw(a, b - 1) % y : pw(a * a % y, b >> 1) : 1;
}
int main() {
  cin >> n >> m >> seed >> maxVal;
  for (int i = 1; i <= n; i++) mp[i] = rnd() % maxVal + 1;
  mp[n + 1] = 0;
  while (m--) {
    op = rnd() % 4 + 1;
    l = rnd() % n + 1;
    r = rnd() % n + 1;
    if (l > r) swap(l, r);
    r++;
    if (op == 3) {
      x = rnd() % (r - l) + 1;
    } else
      x = rnd() % maxVal + 1;
    if (op == 4) y = rnd() % maxVal + 1;
    auto L = --mp.upper_bound(l);
    if (L->first != l) mp[l] = L->second, L++;
    auto R = --mp.upper_bound(r);
    if (R->first != r) mp[r] = R->second, R++;
    if (op == 1) {
      while (L != R) L++->second += x;
    } else if (op == 2) {
      while (L != R) mp.erase(L++), mp[l] = x;
    } else if (op == 3) {
      v.clear();
      for (int d; L != R;)
        d = R--->first, v.push_back({R->second, d - R->first});
      sort(v.begin(), v.end());
      for (auto i : v)
        if (x <= i.second) {
          printf("%lld\n", i.first);
          break;
        } else
          x -= i.second;
    } else {
      long long ans = 0;
      for (int d; L != R;)
        d = R--->first, (ans += pw(R->second % y, x) * (d - R->first)) %= y;
      printf("%lld\n", ans);
    }
  }
}