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;
const long long mx = 2e5 + 5;
long long n, m, vmax;
long long seed, ret, a[mx];
struct Node {
  long long l, r;
  mutable long long val;
  Node(long long a = -1, long long b = -1, long long c = 0) {
    l = a, r = b, val = c;
  }
  bool operator<(const Node &x) const { return l < x.l; }
};
set<Node> s;
long long fpow(long long x, long long y, long long z) {
  long long mul = 1;
  x %= z;
  for (; y; y >>= 1) {
    if (y & 1) mul = mul * x % z;
    x = x * x % z;
  }
  return mul;
}
long long rnd() {
  ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
set<Node>::iterator Split(long long pos) {
  set<Node>::iterator it = s.lower_bound(Node(pos));
  if (it != s.end() && it->l == pos) return it;
  it--;
  Node tmp = *it;
  s.erase(it);
  s.insert(Node(tmp.l, pos - 1, tmp.val));
  return s.insert(Node(pos, tmp.r, tmp.val)).first;
}
void Add(long long l, long long r, long long x) {
  set<Node>::iterator tl = Split(l), tr = Split(r + 1);
  for (set<Node>::iterator it = tl; it != tr; it++) {
    it->val += x;
  }
}
void Assign(long long l, long long r, long long x) {
  set<Node>::iterator tl = Split(l), tr = Split(r + 1);
  s.erase(tl, tr);
  s.insert(Node{l, r, x});
}
long long GetFun(long long l, long long r, long long x, long long y) {
  long long res = 0;
  set<Node>::iterator tl = Split(l), tr = Split(r + 1);
  for (set<Node>::iterator it = tl; it != tr; it++) {
    res = (res + fpow(it->val, x, y) * (it->r - it->l + 1) % y) % y;
  }
  return res;
}
long long GetRank(long long l, long long r, long long x) {
  vector<pair<long long, long long> > vec(0);
  set<Node>::iterator tl = Split(l), tr = Split(r + 1);
  for (set<Node>::iterator it = tl; it != tr; it++) {
    vec.push_back(make_pair(it->val, it->r - it->l + 1));
  }
  sort(vec.begin(), vec.end());
  long long cnt = 0;
  for (long long i = 0; i < vec.size(); i++) {
    cnt += vec[i].second;
    if (cnt >= x) return vec[i].first;
  }
  return -1;
}
signed main() {
  scanf("%lld%lld%lld%lld", &n, &m, &seed, &vmax);
  for (long long i = 1; i <= n; i++) {
    a[i] = (rnd() % vmax) + 1;
    s.insert(Node{i, i, a[i]});
  }
  for (long long i = 1; i <= m; i++) {
    long long op = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1, 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;
    }
    if (op == 1) {
      Add(l, r, x);
    } else if (op == 2) {
      Assign(l, r, x);
    } else if (op == 3) {
      printf("%lld\n", GetRank(l, r, x));
    } else {
      printf("%lld\n", GetFun(l, r, x, y));
    }
  }
}