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;
void ga(int N, int *A) {
  for (int i(0); i < N; i++) scanf("%d", A + i);
}
long long ret, seed, n, m, vmax, a[100005], l, r, op, x, y, DB;
int rnd() {
  ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
set<pair<pair<int, int>, long long> > S;
vector<pair<pair<int, int>, long long> > o, q;
void ASS() {
  int E = 0;
  for (auto &h : S) E += h.first.second - h.first.first + 1;
  assert(E == n);
}
void st(int l, int r) {
  o.clear(), q.clear();
  for (auto &h : S)
    if (h.first.first >= l && h.first.second <= r)
      o.push_back(h);
    else if (max(l, h.first.first) <= min(r, h.first.second))
      q.push_back(h);
  for (auto &h : o) S.erase(h);
  for (auto &h : q) {
    S.erase(h);
    if (h.first.first < l && h.first.second > r)
      S.insert({{h.first.first, l - 1}, h.second}),
          S.insert({{r + 1, h.first.second}, h.second}),
          o.push_back({{l, r}, h.second});
    else if (h.first.first < l)
      S.insert({{h.first.first, l - 1}, h.second}),
          o.push_back({{l, h.first.second}, h.second});
    else
      S.insert({{r + 1, h.first.second}, h.second}),
          o.push_back({{h.first.first, r}, h.second});
  }
}
void add(int l, int r, long long v) {
  st(l, r);
  for (auto &h : o) S.insert({h.first, h.second + v});
}
void st(int l, int r, long long v) {
  st(l, r);
  S.insert({{l, r}, v});
}
long long kth(int l, int r, long long K) {
  vector<pair<long long, long long> > u;
  for (auto &h : S)
    if (max(l, h.first.first) <= min(r, h.first.second))
      u.push_back(
          {h.second, min(r, h.first.second) - max(l, h.first.first) + 1ll});
  sort(u.begin(), u.end());
  for (auto &h : u)
    if (K <= h.second)
      return h.first;
    else
      K -= h.second;
  assert(0);
}
long long pw(long long n, long long k, long long MOD) {
  long long r(1);
  while (k) {
    if (k & 1) r *= n, r %= MOD;
    n *= n, n %= MOD;
    k >>= 1;
  }
  return r;
}
long long pwr(int l, int r, long long x, long long M) {
  long long T = 0;
  for (auto &h : S)
    if (max(l, h.first.first) <= min(r, h.first.second))
      T = (T + (min(r, h.first.second) - max(l, h.first.first) + 1ll) *
                   pw(h.second % M, x, M)) %
          M;
  return T;
}
int main(void) {
  scanf("%lld%lld%lld%lld", &n, &m, &seed, &vmax);
  for (int i(1); i < n + 1; i++) a[i] = (rnd() % vmax) + 1;
  for (int i(0); i < n; i++) S.insert({{i, i}, a[i + 1]});
  for (int i(1); i < m + 1; i++) {
    op = (rnd() % 4) + 1;
    l = (rnd() % n) + 1;
    r = (rnd() % n) + 1;
    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 - 1, r - 1, x);
    else if (op == 2)
      st(l - 1, r - 1, x);
    else if (op == 3)
      printf("%lld\n", kth(l - 1, r - 1, x)), ++DB;
    else
      printf("%lld\n", pwr(l - 1, r - 1, x, y)), ++DB;
  }
  return 0;
}