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>
inline long long pow(long long a, long long b, long long mod) {
  long long ans = 1;
  a %= mod;
  for (; b; b >>= 1, a = a * a % mod)
    if (b & 1) ans = ans * a % mod;
  return ans;
}
struct Node {
  int L, R;
  mutable long long val;
  Node(int L, int R = -1, long long val = 0) : L(L), R(R), val(val) {}
  inline bool operator<(const Node& b) const { return L < b.L; }
};
std::set<Node> Chtholly;
inline std::set<Node>::iterator spilt(int pos) {
  std::set<Node>::iterator iter = Chtholly.lower_bound(Node(pos));
  if (iter != Chtholly.end() && iter->L == pos) return iter;
  --iter;
  int L = iter->L, R = iter->R;
  long long V = iter->val;
  Chtholly.erase(iter);
  Chtholly.insert(Node(L, pos - 1, V));
  return Chtholly.insert(Node(pos, R, V)).first;
}
inline void Add(int L, int R, long long val = 1) {
  std::set<Node>::iterator end = spilt(R + 1), iter = spilt(L);
  for (; iter != end; ++iter) iter->val += val;
}
inline void assign(int L, int R, long long val = 0) {
  std::set<Node>::iterator terR = spilt(R + 1), terL = spilt(L);
  Chtholly.erase(terL, terR);
  Chtholly.insert(Node(L, R, val));
}
inline long long Rank(int L, int R, int k) {
  std::vector<std::pair<long long, int> > s;
  std::set<Node>::iterator end = spilt(R + 1), iter = spilt(L);
  for (; iter != end; ++iter) {
    s.push_back(std::pair<long long, int>(iter->val, iter->R - iter->L + 1));
  }
  std::sort(s.begin(), s.end());
  for (auto iter : s) {
    k -= iter.second;
    if (k <= 0) return iter.first;
  }
  return -1ll;
}
inline long long Sum(int L, int R, long long k, long long mod) {
  std::set<Node>::iterator end = spilt(R + 1), iter = spilt(L);
  long long ans = 0;
  for (; iter != end; ++iter) {
    ans = (ans + 1ll * (iter->R - iter->L + 1) * pow(iter->val, k, mod) % mod) %
          mod;
  }
  return ans;
}
long long n, m, seed, vmax;
const int MOD = 1e9 + 7;
inline long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % MOD;
  return ret;
}
signed main() {
  int i;
  scanf("%lld%lld%lld%lld", &n, &m, &seed, &vmax);
  for (i = 1; i <= n; ++i) Chtholly.insert(Node(i, i, rnd() % vmax + 1));
  Chtholly.insert(Node(n + 1, n + 1, 0));
  for (i = 1; i <= m; ++i) {
    int op = rnd() % 4 + 1, L = rnd() % n + 1, R = rnd() % n + 1, x, y;
    if (L > R) L ^= R ^= L ^= R;
    x = rnd() % (op == 3 ? R - L + 1 : vmax) + 1;
    if (op == 4) y = rnd() % vmax + 1;
    if (op == 1) {
      Add(L, R, x);
    }
    if (op == 2) {
      assign(L, R, x);
    }
    if (op == 3) {
      printf("%lld\n", Rank(L, R, x));
    }
    if (op == 4) {
      printf("%lld\n", Sum(L, R, x, y));
    }
  }
}