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<pair<int, int>, long long> tree;
long long fastPow(long long x, long long y, long long mod) {
  long long r = 1;
  x %= mod;
  for (; y; y >>= 1LL) {
    if (y & 1LL) (r *= x) %= mod;
    (x *= x) %= mod;
  }
  return r;
}
void split(int L) {
  auto it = tree.lower_bound({L, 0});
  if (it != tree.end() && (it == tree.begin() || it->first.first == L)) return;
  --it;
  if (it->first.second < L) return;
  int st = it->first.first, en = it->first.second;
  long long val = it->second;
  tree.erase(it);
  tree[{st, L - 1}] = val;
  tree[{L, en}] = val;
}
void upd(int L, int R, int x) {
  split(L);
  split(R + 1);
  for (auto it = tree.lower_bound({L, -1});
       it != tree.end() && it->first.second <= R; tree.erase(it++))
    ;
  tree[{L, R}] = x;
}
void add(int L, int R, int x) {
  split(L);
  split(R + 1);
  for (auto it = tree.lower_bound({L, -1});
       it != tree.end() && it->first.second <= R; ++it) {
    it->second += x;
  }
}
long long getSum(int L, int R, int x, int y) {
  split(L);
  split(R + 1);
  long long sum = 0;
  for (auto it = tree.lower_bound({L, -1});
       it != tree.end() && it->first.second <= R; ++it) {
    sum = (sum + 1LL * fastPow(it->second, x, y) *
                     (it->first.second - it->first.first + 1) % y) %
          y;
  }
  return sum;
}
long long getKth(int L, int R, int k) {
  split(L);
  split(R + 1);
  vector<pair<long long, int> > v;
  for (auto it = tree.lower_bound({L, -1});
       it != tree.end() && it->first.second <= R; ++it) {
    v.push_back({it->second, (it->first.second) - (it->first.first) + 1});
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); ++i) {
    if (v[i].second >= k)
      return v[i].first;
    else
      k -= v[i].second;
  }
}
long long seed;
int rnd() {
  int ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m, v;
  cin >> n >> m >> seed >> v;
  for (int i = 1; i <= n; ++i) {
    int a = rnd() % v + 1;
    tree[{i, i}] = a;
  }
  while (m--) {
    int op = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1, x, y;
    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;
    if (op == 1)
      add(l, r, x);
    else if (op == 2)
      upd(l, r, x);
    else if (op == 3)
      cout << getKth(l, r, x) << "\n";
    else
      cout << getSum(l, r, x, y) << "\n";
  }
  return 0;
}