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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long int mul(long long int a, long long int b, long long int m) {
  return a * b >= m ? a * b % m : a * b;
}
long long int mod = 1e9 + 7;
long long int mpow(long long int b, long long int exp, long long int m) {
  long long int r = 1;
  while (exp) {
    if (exp & 1) r = mul(r, b, m);
    b = mul(b, b, m);
    exp /= 2;
  }
  return r;
}
long long int seed;
long long int rnd() {
  int res = seed;
  seed = (seed * 7 + 13) % mod;
  return res;
}
int main() {
  map<int, long long int> a;
  int n, m, vmax;
  cin >> n >> m >> seed >> vmax;
  for (int i = 1; i <= n; ++i) a.insert(a.end(), {i, rnd() % vmax + 1});
  a.insert(a.end(), {n + 1, 0});
  while (m--) {
    int op = (rnd() % 4) + 1;
    int l = (rnd() % n) + 1;
    int r = (rnd() % n) + 1;
    long long int x, y;
    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;
    auto itl = --(a.upper_bound(l));
    if (itl->first != l) {
      a[l] = itl->second;
      ++itl;
    }
    auto itr = --(a.upper_bound(r + 1));
    if (itr->first != r + 1) {
      a[r + 1] = itr->second;
      ++itr;
    }
    vector<pair<long long int, long long int> > v;
    long long int res = 0;
    if (op == 1) {
      while (itl != itr) {
        --itr;
        itr->second += x;
      }
    } else if (op == 2) {
      while (itl != itr) {
        a.erase(itl++);
      }
      a[l] = x;
    } else if (op == 3) {
      for (int ub; itl != itr;)
        ub = (itr--)->first, v.push_back({itr->second, ub - itr->first});
      sort(v.begin(), v.end());
      for (int i = 0, cnt = 0;; ++i)
        if (x <= (cnt += v[i].second)) {
          cout << v[i].first << "\n";
          break;
        }
    } else {
      for (long long int ub; itl != itr;) {
        ub = (itr--)->first;
        res += (mpow((itr->second) % y, x, y) * (ub - itr->first)) % y;
        res %= y;
      }
      cout << res << "\n";
    }
  }
  return 0;
}