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;
const int MOD = 1000000007;
const int INF = 1000000001;
const long long LINF = 1000000000000000001LL;
int n, m, seed, vmax;
int rnd() {
  int ret = seed;
  seed = ((long long)seed * 7 + 13) % 1000000007;
  return ret;
}
long long expo(long long a, long long n, long long mod) {
  long long ans = 1;
  while (n) {
    if (n & 1LL) ans = (ans * a) % mod;
    n >>= 1;
    a = (a * a) % mod;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m >> seed >> vmax;
  struct inter {
    int l, r;
    mutable long long x;
    bool operator<(const inter &other) const { return l < other.l; }
  };
  set<inter> s;
  for (int i = (0); i <= (n - 1); ++i) {
    int a = 1 + rnd() % vmax;
    s.insert({i, i, a});
  }
  auto cut = [&](int i) {
    auto it = s.lower_bound({i, -1, -1});
    if (it == s.end() || it->l > i) {
      it--;
      int l = it->l, r = it->r;
      long long x = it->x;
      s.erase(it);
      s.insert({l, i - 1, x});
      s.insert({i, r, x});
    }
  };
  vector<pair<long long, int>> values;
  for (int _ = (1); _ <= (m); ++_) {
    int op = 1 + rnd() % 4;
    int l = rnd() % n;
    int r = rnd() % n;
    int x, y;
    if (l > r) swap(l, r);
    if (op == 3) {
      x = rnd() % (r - l + 1);
    } else {
      x = (rnd() % vmax) + 1;
    }
    if (op == 4) {
      y = (rnd() % vmax) + 1;
    }
    cut(l);
    if (r < n - 1) {
      cut(r + 1);
    }
    auto itL = s.lower_bound({l, -1, -1});
    auto itR = s.upper_bound({r, -1, -1});
    if (op == 1) {
      for (auto it = itL; it != itR; it++) {
        it->x += x;
      }
    } else if (op == 2) {
      int lAll = itL->l, rAll = prev(itR)->r;
      s.erase(itL, itR);
      s.insert({lAll, rAll, x});
    } else if (op == 3) {
      values.clear();
      for (auto it = itL; it != itR; it++) {
        values.push_back({it->x, it->r - it->l + 1});
      }
      sort(values.begin(), values.end());
      for (auto &e : values) {
        if (x < e.second) {
          cout << e.first << '\n';
          break;
        }
        x -= e.second;
      }
    } else {
      long long ans = 0;
      for (auto it = itL; it != itR; it++) {
        long long val = it->x;
        if (val >= y) {
          val %= y;
        }
        val = expo(val, x, y);
        val = (val * (it->r - it->l + 1)) % y;
        ans += val;
        if (ans >= y) {
          ans -= y;
        }
      }
      cout << ans << '\n';
    }
  }
  return 0;
}