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 n, m, seed, vmax;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
long long exp(long long x, long long y, long long m) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return exp(x * x % m, y / 2, m);
  else
    return x * exp(x, y - 1, m) % m;
}
void solve() {
  cin >> n >> m >> seed >> vmax;
  map<int, long long> A;
  vector<int> a(n + 1);
  for (int i = 1; i <= (n); ++i) {
    a[i] = (rnd() % vmax) + 1;
    A.insert({i, a[i]});
  }
  A[n + 1] = 0;
  while (m--) {
    int op = (rnd() % 4) + 1;
    int l = (rnd() % n) + 1;
    int r = (rnd() % n) + 1;
    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 it_l = A.upper_bound(l);
    it_l--;
    if (it_l->first != l) {
      A[l] = it_l->second;
      it_l++;
    }
    auto it_r = A.upper_bound(r + 1);
    it_r--;
    if (it_r->first != r + 1) {
      A[r + 1] = it_r->second;
      it_r++;
    }
    switch (op) {
      case 1:
        while (it_l != it_r) {
          it_l->second += x;
          it_l++;
        }
        break;
      case 2:
        it_l->second = x;
        it_l++;
        while (it_l != it_r) {
          A.erase(it_l++);
        }
        break;
      case 3: {
        vector<pair<long long, int>> v;
        int prev = l;
        long long val = it_l->second;
        it_l++;
        while (it_l != it_r) {
          v.push_back({val, it_l->first - prev});
          val = it_l->second;
          prev = it_l->first;
          it_l++;
        }
        v.push_back({val, it_l->first - prev});
        sort(v.begin(), v.end());
        int cnt = 0;
        for (int i = 0; i <= ((int)(v).size() - 1); ++i) {
          cnt += v[i].second;
          if (cnt >= x) {
            cout << v[i].first << endl;
            break;
          }
        }
      } break;
      case 4: {
        vector<tuple<long long, int>> v;
        int prev = l;
        long long val = it_l->second;
        it_l++;
        while (it_l != it_r) {
          v.push_back({val, it_l->first - prev});
          val = it_l->second;
          prev = it_l->first;
          it_l++;
        }
        v.push_back({val, it_l->first - prev});
        long long ans = 0;
        for (int i = 0; i <= ((int)(v).size() - 1); ++i) {
          ans += ((long long)exp(get<0>(v[i]) % y, x, y) * get<1>(v[i]) % y);
          ans %= y;
        }
        cout << ans << endl;
        break;
      }
    }
  }
}
int main() {
  clock_t beg = clock();
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  clock_t end = clock();
  fprintf(stderr, "%lf\n", (double)(end - beg) / CLOCKS_PER_SEC);
}