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;
struct Node {
  long long l, r;
  mutable long long val;
  Node(long long l = 0, long long r = 0, long long v = 0)
      : l(l), r(r), val(v) {}
  bool operator<(const Node& rv) const { return l < rv.l; }
};
long long n, m;
set<Node> t;
long long a[100005];
long long seed;
long long vmax;
long long qpow(long long x, long long p, long long m) {
  long long ans = 1;
  long long base = x;
  while (p) {
    if (p & 1) ans = (ans * base) % m;
    base = (base * base) % m;
    p >>= 1;
  }
  return ans;
}
long long rnd() {
  long long ans = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ans;
}
set<Node>::iterator split(long long p) {
  set<Node>::iterator it = t.lower_bound(Node(p));
  if (it != t.end() && it->l == p) {
    return it;
  }
  --it;
  long long l = it->l, r = it->r, val = it->val;
  t.erase(it);
  t.insert(Node(l, p - 1, val));
  return t.insert(Node(p, r, val)).first;
}
void assign(long long l, long long r, long long v) {
  set<Node>::iterator end = split(r + 1);
  set<Node>::iterator begin = split(l);
  t.erase(begin, end);
  t.insert(Node(l, r, v));
}
void add(long long l, long long r, long long k) {
  set<Node>::iterator end = split(r + 1);
  set<Node>::iterator begin = split(l);
  for (set<Node>::iterator it = begin; it != end; ++it) {
    it->val += k;
  }
}
long long kth(long long l, long long r, long long k) {
  vector<pair<long long, long long> > v;
  set<Node>::iterator end = split(r + 1);
  set<Node>::iterator begin = split(l);
  for (set<Node>::iterator it = begin; it != end; ++it) {
    v.push_back(make_pair(it->val, it->r - it->l + 1));
  }
  sort(v.begin(), v.end());
  for (long long i = 0; i < v.size(); ++i) {
    k -= v[i].second;
    if (k <= 0) return v[i].first;
  }
  return -1;
}
long long power(long long l, long long r, long long k, long long mod) {
  set<Node>::iterator end = split(r + 1);
  set<Node>::iterator begin = split(l);
  long long ans = 0;
  for (set<Node>::iterator it = begin; it != end; ++it) {
    ans = (ans +
           (((qpow(it->val % mod, k, mod) % mod) * (it->r - it->l + 1 % mod)) %
            mod)) %
          mod;
  }
  return ans % mod;
}
signed main() {
  ios::sync_with_stdio(false);
  cin >> n >> m >> seed >> vmax;
  for (long long i = 1; i <= n; ++i) {
    a[i] = (rnd() % vmax) + 1;
    t.insert(Node(i, i, a[i]));
  }
  for (long long i = 1; i <= m; ++i) {
    long long opt = (rnd() % 4) + 1, l = (rnd() % n) + 1, r = (rnd() % n) + 1,
              x, y;
    if (l > r) swap(l, r);
    if (opt == 3)
      x = (rnd() % (r - l + 1)) + 1;
    else
      x = (rnd() % vmax) + 1;
    if (opt == 4) y = (rnd() % vmax) + 1;
    if (opt == 1) {
      add(l, r, x);
    } else if (opt == 2) {
      assign(l, r, x);
    } else if (opt == 3) {
      cout << kth(l, r, x) << endl;
    } else {
      cout << power(l, r, x, y) << endl;
    }
  }
  return 0;
}