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>
#pragma GCC optimize(2)
using namespace std;
const int maxn = 100005;
inline int rnd(long long &seed) {
  register int ans = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ans;
}
map<int, long long> ct;
pair<long long, int> cv[maxn];
inline long long mypow(long long base, int x, int mod) {
  register long long ans = 1;
  base %= mod;
  while (true) {
    if (x & 1) ans = ans * base % mod;
    base = base * base % mod;
    if ((x >>= 1) == 0) return ans;
  }
}
int main() {
  int n, m, vmax;
  long long seed;
  scanf("%d%d%lld%d", &n, &m, &seed, &vmax);
  for (int i = 0; i < n; i++) ct[i] = (rnd(seed) % vmax) + 1;
  ct[n] = 0;
  while (m--) {
    register int op = rnd(seed) % 4;
    register int l = rnd(seed) % n;
    register int r = rnd(seed) % n;
    register int tot = 0;
    if (l > r) swap(l, r);
    r++;
    register int x = (rnd(seed) % ((op != 2) ? vmax : (r - l))) + 1, y;
    if (op == 3) y = (rnd(seed) % vmax) + 1;
    register auto itl = ct.lower_bound(l);
    if (itl->first != l) {
      itl--;
      ct.insert(itl, {l, itl->second});
      itl++;
    }
    register auto itr = ct.lower_bound(r);
    if (itr->first != r) {
      itr--;
      ct.insert(itr, {r, itr->second});
      itr++;
    }
    switch (op) {
      case 0:
        while (itl != itr) (itl++)->second += x;
        break;
      case 1:
        while (itl != itr) ct.erase(itl++);
        ct.insert(prev(itr), {l, x});
        break;
      case 2:
        do {
          itr--;
          cv[tot++] = {itr->second, r - itr->first};
          r = itr->first;
        } while (itl != itr);
        sort(cv, &cv[tot]);
        l = -1;
        r = 0;
        do {
          r += cv[++l].second;
        } while (r < x);
        printf("%lld\n", cv[l].first);
        break;
      case 3:
        register long long ans = 0;
        do {
          itr--;
          ans = (ans + (r - itr->first) * mypow(itr->second, x, y)) % y;
          r = itr->first;
        } while (itr != itl);
        printf("%lld\n", ans);
        break;
    }
  }
  return 0;
}