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;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long poww(long long a, long long b) {
  if (b == 0) return 1;
  long long tmp = poww(a, b / 2);
  return (b & 1 ? a * tmp * tmp : tmp * tmp);
}
string itos(long long i) {
  string s = "";
  while (i) {
    s += char(i % 10 + '0');
    i /= 10;
  }
  reverse(s.begin(), s.end());
  return s;
}
long long stoi(string &s) {
  long long tot = 0;
  for (int i = (int)s.length() - 1, j = 1; i >= 0; i--, j *= 10) {
    tot += j * (s[i] + '0');
  }
  return tot;
}
int months[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
using namespace std;
void tt() { freopen("test.txt", "r", stdin); }
const int MAX = 1e6 + 10;
int n, m, seed, vmax;
int a[MAX];
int rnd() {
  int ret = seed;
  seed = (1ll * seed * 7 + 13) % (1000 * 1000 * 1000 + 7);
  return ret;
  ;
}
long long pw(long long base, long long ex, long long mod) {
  if (ex == 0) return 1 % mod;
  long long res = pw(base, ex / 2, mod);
  ;
  res = (res * res) % mod;
  if (ex % 2 == 1) res = (res * (base % mod)) % mod;
  return res;
}
map<int, long long> s;
int main() {
  cin >> n >> m >> seed >> vmax;
  for (int i = 1; i <= n; i++) a[i] = (rnd() % vmax) + 1;
  for (int i = 1; i <= n; i++) s.insert({i, a[i]});
  for (int i = 1; i <= m; i++) {
    int x, y;
    int op = (rnd() % 4) + 1;
    int l = (rnd() % n) + 1;
    int 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;
    auto itl = s.upper_bound(l);
    --itl;
    if (itl->first != l) {
      s[l] = itl->second;
      ++itl;
    }
    auto itr = s.upper_bound(r + 1);
    --itr;
    if (itr->first != r + 1) {
      s[r + 1] = itr->second;
      ++itr;
    }
    if (op == 1) {
      while (itl != itr) {
        itl->second += x;
        itl++;
      }
    } else if (op == 2) {
      while (itl != itr) {
        s.erase(itl++);
      }
      s[l] = x;
    } else if (op == 3) {
      vector<pair<long long, int> > v;
      while (itr != itl) {
        int a = itr->first;
        itr--;
        int b = itr->first;
        v.push_back({itr->second, a - b});
      }
      long long res = 0, cnt = 0;
      sort(v.begin(), v.end());
      for (auto t : v) {
        cnt += t.second;
        if (cnt >= x) {
          res = t.first;
          break;
        }
      }
      cout << res << '\n';
    } else if (op == 4) {
      vector<pair<long long, int> > v;
      while (itr != itl) {
        int a = itr->first;
        itr--;
        int b = itr->first;
        v.push_back({itr->second, a - b});
      }
      long long res = 0;
      for (auto t : v) {
        res = res + 1ll * pw(t.first, x, y) % y * t.second % y;
        res %= y;
      }
      cout << res << '\n';
    }
  }
  return 0;
}