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 read() {
  long long x = 0;
  int zf = 1;
  char ch = ' ';
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') zf = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * zf;
}
namespace Qpow {
long long pow(long long a, long long b, long long mod) {
  if (!a) return 0;
  long long res = 1;
  a %= mod;
  for (; b; (a *= a) %= mod, b >>= 1ll)
    if (b & 1) (res *= a) %= mod;
  ;
  return res;
}
};  // namespace Qpow
struct Node {
  int l, r;
  mutable long long val;
  Node(int a = -1, int b = -1, long long c = 0) { l = a, r = b, val = c; }
  bool operator<(const Node &a) const { return l < a.l; }
};
set<Node> st;
set<Node>::iterator split(int pos) {
  set<Node>::iterator it = st.lower_bound(Node(pos));
  if (it != st.end() && it->l == pos) return it;
  --it;
  Node tmp = *it;
  st.erase(it);
  st.insert(Node(tmp.l, pos - 1, tmp.val));
  return st.insert(Node(pos, tmp.r, tmp.val)).first;
}
void assign(int l, int r, long long val) {
  set<Node>::iterator itr = split(r + 1), itl = split(l);
  st.erase(itl, itr);
  st.insert((Node){l, r, val});
}
void add(int l, int r, long long val) {
  set<Node>::iterator itr = split(r + 1), itl = split(l);
  for (set<Node>::iterator it = itl; it != itr; ++it) it->val += val;
}
long long querySum(int l, int r) {
  set<Node>::iterator itr = split(r + 1), itl = split(l);
  long long res = 0;
  for (set<Node>::iterator it = itl; it != itr; ++it)
    res += (it->r - it->l + 1) * it->val;
  return res;
}
long long querySumWithPow(int l, int r, long long x, long long mod) {
  set<Node>::iterator itr = split(r + 1), itl = split(l);
  long long res = 0;
  for (set<Node>::iterator it = itl; it != itr; ++it)
    (res += (it->r - it->l + 1) * Qpow::pow(it->val, x, mod)) %= mod;
  return res;
}
long long queryKth(int l, int r, int k) {
  vector<pair<long long, int> > vec(0);
  set<Node>::iterator itr = split(r + 1), itl = split(l);
  for (set<Node>::iterator it = itl; it != itr; ++it)
    vec.push_back(make_pair(it->val, it->r - it->l + 1));
  sort(vec.begin(), vec.end());
  for (vector<pair<long long, int> >::iterator it = vec.begin();
       it != vec.end(); ++it)
    if ((k -= it->second) <= 0) return it->first;
  return -1;
}
long long seed;
long long a[100005];
inline long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
int main() {
  int n = read(), m = read();
  seed = read();
  long long vmax = read();
  for (int i = 1; i <= n; ++i) {
    a[i] = (rnd() % vmax) + 1;
    st.insert((Node){i, i, a[i]});
  }
  long long x, y;
  for (int i = 1; i <= m; ++i) {
    int op = (rnd() % 4) + 1, l = (rnd() % n) + 1, r = (rnd() % n) + 1;
    if (l > r) {
      int tmp = l;
      l = r, r = tmp;
    }
    if (op == 3)
      x = (rnd() % (r - l + 1)) + 1;
    else
      x = (rnd() % vmax) + 1;
    if (op == 4) y = (rnd() % vmax) + 1;
    if (op == 1)
      add(l, r, x);
    else if (op == 2)
      assign(l, r, x);
    else if (op == 3)
      printf("%I64d\n", queryKth(l, r, x));
    else if (op == 4)
      printf("%I64d\n", querySumWithPow(l, r, x, y));
  }
  return 0;
}