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 GCC optimize("Ofast", "unroll-loops")
using namespace std;
const long long inf = numeric_limits<long long>::max() / 2;
const long double eps = 1e-9;
const long double pi = acos(-1);
inline void solve(), read();
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  read();
  solve();
  return 0;
}
long long n, m, seed, vmax;
map<long long, long long> q;
long long binpow(long long n, long long p, long long r) {
  if (p == 0) {
    return 1 % r;
  }
  if (p == 1) {
    return n % r;
  }
  long long ans = binpow(n, p / 2, r);
  ans = ans * ans % r;
  if (p % 2 == 1) {
    ans = ans * n % r;
  }
  return ans;
}
long long rnd() {
  long long ans = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ans;
}
inline void solve() {
  for (long long i = 1; i <= n; i++) {
    q[i] = rnd() % vmax + 1;
  }
  q[n + 1] = 0;
  for (long long i = 0; i < m; i++) {
    long long op = rnd() % 4 + 1;
    long long l = rnd() % n + 1;
    long long r = rnd() % n + 1;
    long long x = 0, y = 0;
    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 = q.upper_bound(l);
    --itl;
    if (itl->first != l) {
      q[l] = itl->second;
      ++itl;
    }
    auto itr = q.upper_bound(r + 1);
    --itr;
    if (itr->first != r + 1) {
      q[r + 1] = itr->second;
      ++itr;
    }
    if (op == 1) {
      for (auto& iter = itl; iter != itr; iter++) {
        iter->second += x;
      }
    } else if (op == 2) {
      for (auto& iter = itl; iter != itr;) {
        iter = q.erase(iter);
      }
      q[l] = x;
    } else if (op == 3) {
      vector<pair<long long, long long> > v;
      for (auto& iter = itl; iter != itr; iter++) {
        long long n = next(iter)->first - iter->first;
        v.push_back({iter->second, n});
      }
      long long cur = 0;
      sort((v).begin(), (v).end());
      for (pair<long long, long long> p : v) {
        if (cur + p.second >= x) {
          cout << p.first << "\n";
          break;
        }
        cur += p.second;
      }
    } else if (op == 4) {
      long long ans = 0;
      for (auto& iter = itl; iter != itr; iter++) {
        long long n = next(iter)->first - iter->first;
        long long val = binpow(iter->second % y, x, y);
        ans += n % y * val % y;
        ans %= y;
      }
      cout << ans << "\n";
    }
  }
}
inline void read() { cin >> n >> m >> seed >> vmax; }