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;
struct Node {
  long long l, r;
  mutable long long v;
  Node() {}
  Node(long long a, long long b, long long c) {
    l = a;
    r = b;
    v = c;
  }
};
bool operator<(Node a, Node b) { return a.l < b.l; }
set<Node> st;
set<Node>::iterator split(long long x) {
  set<Node>::iterator tmp = st.lower_bound(Node(x, 0, 0));
  if (tmp != st.end() && tmp->l == x) return tmp;
  tmp--;
  long long l = tmp->l;
  long long r = tmp->r;
  long long v = tmp->v;
  st.erase(tmp);
  st.insert(Node(l, x - 1, v));
  return st.insert(Node(x, r, v)).first;
}
void change(long long l, long long r, long long v) {
  set<Node>::iterator tr = split(r + 1);
  set<Node>::iterator tl = split(l);
  st.erase(tl, tr);
  st.insert(Node(l, r, v));
}
long long kth(long long l, long long r, long long k) {
  vector<pair<long long, long long> > tmp;
  set<Node>::iterator tr = split(r + 1);
  set<Node>::iterator tl = split(l);
  for (set<Node>::iterator i = tl; i != tr; i++)
    tmp.push_back(make_pair(i->v, i->r - i->l + 1));
  sort(tmp.begin(), tmp.end());
  for (long long i = 0; i < tmp.size(); i++) {
    if (k <= tmp[i].second) return tmp[i].first;
    k -= tmp[i].second;
  }
  return 0;
}
void add(long long l, long long r, long long v) {
  set<Node>::iterator tr = split(r + 1);
  set<Node>::iterator tl = split(l);
  for (set<Node>::iterator i = tl; i != tr; i++) {
    i->v += v;
  }
}
long long qpow(long long a, long long b, long long mod) {
  long long tmp = a % mod;
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * tmp % mod;
    tmp = tmp * tmp % mod;
    b >>= 1;
  }
  return ans;
}
long long powsum(long long l, long long r, long long a, long long b) {
  set<Node>::iterator tr = split(r + 1);
  set<Node>::iterator tl = split(l);
  long long ans = 0;
  for (set<Node>::iterator i = tl; i != tr; i++) {
    ans = (ans + qpow(i->v, a, b) * (i->r - i->l + 1) % b) % b;
  }
  return ans;
}
long long n, m, vmax, seed;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m >> seed >> vmax;
  for (long long i = 1; i <= n; i++) {
    st.insert(Node(i, i, rnd() % vmax + 1));
  }
  st.insert(Node(n + 1, n + 1, 0));
  for (long long i = 1; i <= m; i++) {
    long long op = (rnd() % 4) + 1;
    long long l = (rnd() % n) + 1;
    long long r = (rnd() % n) + 1;
    long long x;
    long long y;
    if (l > r) swap(l, r);
    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)
      change(l, r, x);
    else if (op == 3)
      cout << kth(l, r, x) << endl;
    else
      cout << powsum(l, r, x, y) << endl;
  }
  return 0;
}