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 f = 1, x = 0;
  char ss = getchar();
  while (ss < '0' || ss > '9') {
    if (ss == '-') f = -1;
    ss = getchar();
  }
  while (ss >= '0' && ss <= '9') {
    x = x * 10 + ss - '0';
    ss = getchar();
  }
  return f * x;
}
long long seed, vmax;
long long rnd() {
  long long res = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return res;
}
const int maxn = 100010;
int n, m;
long long a[maxn];
struct node {
  int ll, rr;
  mutable long long val;
  node(int L, int R = -1, long long V = 0) : ll(L), rr(R), val(V) {}
  bool operator<(const node& tt) const { return ll < tt.ll; }
};
set<node> st;
long long qpow(long long a, long long k, long long p) {
  long long res = 1;
  a %= p;
  while (k > 0) {
    if (k & 1) res = (res * a) % p;
    a = (a * a) % p;
    k >>= 1;
  }
  return res;
}
set<node>::iterator split(int pos) {
  set<node>::iterator it = st.lower_bound(node(pos));
  if (it != st.end() && it->ll == pos) return it;
  --it;
  int ll = it->ll, rr = it->rr;
  long long val = it->val;
  st.erase(it);
  st.insert(node(ll, pos - 1, val));
  return st.insert(node(pos, rr, val)).first;
}
void assign(int ll, int rr, long long val) {
  set<node>::iterator itr = split(rr + 1), itl = split(ll);
  st.erase(itl, itr);
  st.insert(node(ll, rr, val));
}
void add(int ll, int rr, long long val) {
  set<node>::iterator itr = split(rr + 1), itl = split(ll);
  for (; itl != itr; ++itl) itl->val += val;
}
long long kth(int ll, int rr, int k) {
  vector<pair<long long, int> > vec;
  set<node>::iterator itr = split(rr + 1), itl = split(ll);
  for (; itl != itr; ++itl)
    vec.push_back(pair<long long, int>(itl->val, itl->rr - itl->ll + 1));
  sort(vec.begin(), vec.end());
  for (vector<pair<long long, int> >::iterator it = vec.begin();
       it != vec.end(); ++it) {
    k -= it->second;
    if (k <= 0) return it->first;
  }
  return -1;
}
long long qsum(int ll, int rr, long long x, long long y) {
  long long res = 0;
  set<node>::iterator itr = split(rr + 1), itl = split(ll);
  for (; itl != itr; ++itl)
    res += (qpow(itl->val, x, y) * ((itl->rr - itl->ll + 1) % y)) % y, res %= y;
  return res;
}
int main() {
  n = read();
  m = read();
  seed = read();
  vmax = read();
  for (int i = 1; i <= n; ++i) {
    a[i] = (rnd() % vmax) + 1;
    st.insert(node(i, i, a[i]));
  }
  for (int i = 1; i <= m; ++i) {
    int op = (rnd() % 4) + 1;
    int ll = (rnd() % n) + 1, rr = (rnd() % n) + 1;
    long long x, y;
    if (ll > rr) swap(ll, rr);
    if (op == 3)
      x = (rnd() % (rr - ll + 1)) + 1;
    else
      x = (rnd() % vmax) + 1;
    if (op == 4) y = (rnd() % vmax) + 1;
    if (op == 1)
      add(ll, rr, x);
    else if (op == 2)
      assign(ll, rr, x);
    else if (op == 3)
      printf("%lld\n", kth(ll, rr, x));
    else if (op == 4)
      printf("%lld\n", qsum(ll, rr, x, y));
  }
  return 0;
}