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 seed;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
long long modpow(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      ans *= a;
      ans %= mod;
    }
    a *= a;
    a %= mod;
    b /= 2;
  }
  return ans;
}
int main() {
  int n, m, vmax;
  cin >> n >> m >> seed >> vmax;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) {
    a[i] = rnd() % vmax + 1;
  }
  vector<tuple<long long, int, int>> T(n);
  for (int i = 0; i < n; i++) {
    T[i] = make_tuple(a[i], i, i + 1);
  }
  for (int i = 0; i < m; i++) {
    int t = rnd() % 4 + 1;
    int l = rnd() % n + 1;
    int r = rnd() % n + 1;
    if (l > r) {
      swap(l, r);
    }
    l--;
    if (t == 1 || t == 2) {
      int x = rnd() % vmax + 1;
      vector<tuple<long long, int, int>> T2;
      int cnt = T.size();
      for (int j = 0; j < cnt; j++) {
        long long A = get<0>(T[j]);
        int L = get<1>(T[j]);
        int R = get<2>(T[j]);
        if (L < l) {
          T2.push_back(make_tuple(A, L, min(l, R)));
        }
        if (max(L, l) < min(R, r)) {
          if (t == 1) {
            T2.push_back(make_tuple(A + x, max(L, l), min(R, r)));
          } else {
            T2.push_back(make_tuple(x, max(L, l), min(R, r)));
          }
        }
        if (r < R) {
          T2.push_back(make_tuple(A, max(L, r), R));
        }
      }
      int cnt2 = T2.size();
      vector<tuple<long long, int, int>> T3;
      T3.push_back(T2[0]);
      for (int j = 1; j < cnt2; j++) {
        if (get<0>(T2[j - 1]) == get<0>(T2[j])) {
          get<2>(T3.back()) = get<2>(T2[j]);
        } else {
          T3.push_back(T2[j]);
        }
      }
      swap(T, T3);
    }
    if (t == 3) {
      int x = rnd() % (r - l);
      int cnt = T.size();
      vector<pair<long long, int>> P;
      for (int j = 0; j < cnt; j++) {
        long long A = get<0>(T[j]);
        int L = get<1>(T[j]);
        int R = get<2>(T[j]);
        int d = min(R, r) - max(L, l);
        if (d > 0) {
          P.push_back(make_pair(A, d));
        }
      }
      sort(P.begin(), P.end());
      int cnt2 = P.size();
      for (int j = 0; j < cnt2; j++) {
        if (x >= P[j].second) {
          x -= P[j].second;
        } else {
          cout << P[j].first << "\n";
          break;
        }
      }
    }
    if (t == 4) {
      int x = rnd() % vmax + 1;
      int y = rnd() % vmax + 1;
      int cnt = T.size();
      long long ans = 0;
      vector<pair<long long, int>> P;
      for (int j = 0; j < cnt; j++) {
        long long A = get<0>(T[j]);
        int L = get<1>(T[j]);
        int R = get<2>(T[j]);
        int d = min(R, r) - max(L, l);
        if (d > 0) {
          ans += modpow(A % y, x, y) * d;
          ans %= y;
        }
      }
      cout << ans << "\n";
    }
  }
}