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;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m;
long long seed, vmax;
long long rd() {
  long long ret = seed;
  seed = ((long long)seed * 7 + 13) % 1000000007;
  return ret;
}
long long a[100010];
long long quickmod(long long a, long long b, long long md) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % md;
    a = a * a % md;
    b >>= 1;
  }
  return ret;
}
namespace seg {
long long mx[100010 << 2], mn[100010 << 2], tag[100010 << 2];
void pushup(int x) {
  mx[x] = max(mx[x << 1], mx[x << 1 | 1]);
  mn[x] = min(mn[x << 1], mn[x << 1 | 1]);
}
void cv(int x, long long y) { mx[x] = mn[x] = tag[x] = y; }
void build(int x, int l, int r) {
  if (l == r) {
    mn[x] = mx[x] = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(x << 1, l, mid);
  build(x << 1 | 1, mid + 1, r);
  pushup(x);
}
void pushdown(int x) {
  if (tag[x]) {
    mn[x << 1] = mx[x << 1] = tag[x << 1] = tag[x];
    mn[x << 1 | 1] = mx[x << 1 | 1] = tag[x << 1 | 1] = tag[x];
    tag[x] = 0;
  }
}
void add(int x, int l, int r, int L, int R, long long v) {
  if (l >= L && r <= R && mn[x] == mx[x]) {
    cv(x, mx[x] + v);
    return;
  }
  int mid = (l + r) >> 1;
  pushdown(x);
  if (L <= mid) add(x << 1, l, mid, L, R, v);
  if (R > mid) add(x << 1 | 1, mid + 1, r, L, R, v);
  pushup(x);
}
void cov(int x, int l, int r, int L, int R, long long v) {
  if (l >= L && r <= R) {
    cv(x, v);
    return;
  }
  int mid = (l + r) >> 1;
  pushdown(x);
  if (L <= mid) cov(x << 1, l, mid, L, R, v);
  if (R > mid) cov(x << 1 | 1, mid + 1, r, L, R, v);
  pushup(x);
}
struct zgz {
  long long x, y;
} t[100010];
bool cmp(const zgz &a, const zgz &b) { return a.x < b.x; }
int top;
void get(int x, int l, int r, int L, int R) {
  if (l >= L && r <= R && mn[x] == mx[x]) {
    t[++top].x = mn[x];
    t[top].y = r - l + 1;
    return;
  }
  int mid = (l + r) >> 1;
  pushdown(x);
  if (L <= mid) get(x << 1, l, mid, L, R);
  if (R > mid) get(x << 1 | 1, mid + 1, r, L, R);
}
long long ask1(int x, long long y, long long k) {
  top = 0;
  get(1, 1, n, x, y);
  sort(t + 1, t + top + 1, cmp);
  for (int i = 1; i <= top; i++) {
    if (k <= t[i].y) return t[i].x;
    k -= t[i].y;
  }
}
long long ask2(long long l, long long r, long long x, long long y) {
  top = 0;
  get(1, 1, n, l, r);
  sort(t + 1, t + top + 1, cmp);
  long long tx = 0, ty = 0, ret = 0;
  for (int i = 1; i <= top; i++) {
    if (t[i].x != tx) {
      ret += ty * quickmod(tx % y, x, y) % y;
      ret %= y;
      tx = t[i].x;
      ty = 0;
    }
    ty += t[i].y;
  }
  ret += ty * quickmod(tx % y, x, y) % y;
  ret %= y;
  return ret;
}
}  // namespace seg
int main() {
  n = read(), m = read(), seed = read(), vmax = read();
  for (int i = 1; i <= n; i++) a[i] = (rd() % vmax) + 1;
  seg::build(1, 1, n);
  while (m--) {
    int opt = rd() % 4 + 1;
    int l = rd() % n + 1, r = rd() % n + 1;
    if (l > r) swap(l, r);
    if (opt == 1) {
      long long x = (rd() % vmax) + 1;
      seg::add(1, 1, n, l, r, x);
    } else if (opt == 2) {
      long long x = (rd() % vmax) + 1;
      seg::cov(1, 1, n, l, r, x);
    } else if (opt == 3) {
      long long x = (rd() % (r - l + 1)) + 1;
      printf("%lld\n", seg::ask1(l, r, x));
    } else {
      long long x = (rd() % vmax) + 1, y = (rd() % vmax) + 1;
      printf("%lld\n", seg::ask2(l, r, x, y));
    }
  }
}