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;
template <typename A, typename B>
ostream& operator<<(ostream& os, const pair<A, B>& p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream& operator<<(ostream& os, const T_container& v) {
  string sep;
  for (const T& x : v) os << sep << x, sep = " ";
  return os;
}
const int mxN = 1e5 + 1, oo = 1e9;
long long seed;
int n, m, vmax;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
long long MOD = 1e9 + 7;
class mint {
 public:
  int d;
  mint() { d = 0; }
  mint(long long _d) : d(_d % MOD){};
  mint operator*(const mint& o) const { return ((long long)d * o.d) % MOD; }
  mint operator+(const mint& o) const {
    long long tmp = d + o.d;
    if (tmp >= MOD) tmp -= MOD;
    return tmp;
  }
  mint operator-(const mint& o) const {
    long long tmp = d - o.d;
    if (tmp < 0) tmp += MOD;
    return tmp;
  }
  mint operator^(long long b) const {
    mint tmp = 1;
    mint power = *this;
    while (b) {
      if (b & 1) {
        tmp = tmp * power;
      }
      power = power * power;
      b /= 2;
    }
    return tmp;
  }
  mint operator/(const mint& o) { return *this * (o ^ (MOD - 2)); }
  bool operator==(const mint& o) { return d == o.d; }
  friend mint& operator+=(mint& a, const mint& o) {
    a.d += o.d;
    if (a.d >= MOD) a.d -= MOD;
    return a;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m >> seed >> vmax;
  struct iv {
    long long v;
    int r;
  };
  map<int, long long> mp;
  for (int i = 1; i <= n; ++i) {
    mp[i] = (rnd() % vmax) + 1;
  }
  mp[n + 1] = 0;
  for (int i = 1; i <= m; ++i) {
    int op = (rnd() % 4) + 1;
    int l = (rnd() % n) + 1;
    int r = (rnd() % n) + 1;
    if (l > r) swap(l, r);
    int x, y = -1;
    if (op == 3)
      x = (rnd() % (r - l + 1)) + 1;
    else
      x = (rnd() % vmax) + 1;
    if (op == 4) y = (rnd() % vmax) + 1;
    auto cut = [&](int k) {
      auto f = mp.lower_bound(k);
      if (f == mp.end() or f->first != k) {
        auto p = prev(f);
        mp[k] = p->second;
      }
    };
    cut(l);
    cut(r + 1);
    auto a = mp.find(l), b = mp.find(r + 1);
    if (op == 1) {
      while (a != b) {
        a->second += x;
        a++;
      }
    } else if (op == 2) {
      auto beg = a;
      ++a;
      while (a != b) {
        a = mp.erase(a);
      }
      beg->second = x;
    } else if (op == 3) {
      vector<pair<long long, int>> v;
      while (a != b) {
        v.push_back({a->second, next(a)->first - a->first});
        ++a;
      }
      sort(begin(v), end(v));
      int cnt = 0;
      for (auto& [val, times] : v) {
        cnt += times;
        if (cnt >= x) {
          cout << val << '\n';
          break;
        }
      }
    } else {
      MOD = y;
      mint ans = 0;
      while (a != b) {
        ans += (mint(a->second) ^ x) * (next(a)->first - a->first);
        a++;
      }
      cout << ans.d << '\n';
    }
  }
}