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;
long long power(long long base, long long expo, long long MOD) {
  long long res = 1 % MOD;
  while (expo > 0) {
    if (expo & 1) res = (res * base) % MOD;
    base = (base * base) % MOD;
    expo >>= 1;
  }
  return res;
}
long long seed;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
int main() {
  {
    int n, m;
    long long vmax;
    scanf("%d", &n), scanf("%d", &m), scanf("%lld", &seed),
        scanf("%lld", &vmax);
    long long arr;
    map<int, long long> mp;
    for (int i = 0; i < n; ++i) {
      arr = (rnd() % vmax) + 1;
      mp.insert(mp.end(), {i + 1, arr});
    }
    mp.insert(mp.end(), {n + 1, 0LL});
    for (int j = 0; j < m; ++j) {
      int op, l, r;
      op = (rnd() % 4) + 1;
      l = (rnd() % n) + 1;
      r = (rnd() % n) + 1;
      if (l > r) swap(l, r);
      long long x, y;
      auto leftIterator = mp.lower_bound(l);
      if (leftIterator->first != l)
        --leftIterator, mp[l] = (leftIterator)->second,
                        leftIterator = mp.lower_bound(l);
      auto iteratorNextToRight = mp.lower_bound(r + 1);
      if (iteratorNextToRight->first != r + 1)
        --iteratorNextToRight,
            mp[r + 1] = iteratorNextToRight->second,
                   iteratorNextToRight = mp.lower_bound(r + 1);
      switch (op) {
        case 1: {
          x = (rnd() % vmax) + 1;
          while (leftIterator != iteratorNextToRight) {
            leftIterator->second += x;
            leftIterator++;
          }
          break;
        }
        case 2: {
          x = (rnd() % vmax) + 1;
          mp.erase(leftIterator, iteratorNextToRight);
          mp[l] = x;
          break;
        }
        case 3: {
          x = (rnd() % (r - l + 1)) + 1;
          vector<pair<long long, long long>> sol;
          for (; leftIterator != iteratorNextToRight;) {
            long long val = leftIterator->second;
            int l = leftIterator->first;
            ++leftIterator;
            int cnt = min((leftIterator)->first, r + 1) - l;
            sol.push_back({val, cnt});
          }
          sort(sol.begin(), sol.end());
          int cnt = 0;
          for (int i = 0; i < sol.size(); ++i) {
            cnt += sol[i].second;
            if (cnt >= x) {
              printf("%lld\n", sol[i]);
              break;
            }
          }
          break;
        }
        case 4: {
          x = (rnd() % vmax) + 1;
          y = (rnd() % vmax) + 1;
          long long ans = 0;
          for (; leftIterator != iteratorNextToRight;) {
            long long val = leftIterator->second;
            int l = leftIterator->first;
            ++leftIterator;
            int cnt = min((leftIterator)->first, r + 1) - l;
            ans = (ans + (power(val % y, x, y) * cnt) % y) % y;
          }
          printf("%lld\n", ans);
          break;
        }
      }
    }
  }
  return 0;
}