MLE 应该是数组开多了导致的。

最后统计用快速幂没有 TLE 比较神秘。

但是 WA 是啥呢。。。

#include <bits/stdc++.h>
using lint = long long;
const int N = 5e6 + 10;
#define int long long

int n, k, p;
lint ans, a[N], s[N], t[N], inv[N];

// #define DEBUG 1  // 调试开关
struct IO {
#define MAXSIZE (1 << 20)
#define isdigit(x) (x >= '0' && x <= '9')
  char buf[MAXSIZE], *p1, *p2;
  char pbuf[MAXSIZE], *pp;
#if DEBUG
#else
  IO() : p1(buf), p2(buf), pp(pbuf) {}

  ~IO() { fwrite(pbuf, 1, pp - pbuf, stdout); }
#endif
  char gc() {
#if DEBUG  // 调试，可显示字符
    return getchar();
#endif
    if (p1 == p2) p2 = (p1 = buf) + fread(buf, 1, MAXSIZE, stdin);
    return p1 == p2 ? ' ' : *p1++;
  }

  bool blank(char ch) {
    return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
  }

  template <class T>
  void read(T &x) {
    double tmp = 1;
    bool sign = 0;
    x = 0;
    char ch = gc();
    for (; !isdigit(ch); ch = gc())
      if (ch == '-') sign = 1;
    for (; isdigit(ch); ch = gc()) x = x * 10 + (ch - '0');
    if (ch == '.')
      for (ch = gc(); isdigit(ch); ch = gc())
        tmp /= 10.0, x += tmp * (ch - '0');
    if (sign) x = -x;
  }

  void read(char *s) {
    char ch = gc();
    for (; blank(ch); ch = gc());
    for (; !blank(ch); ch = gc()) *s++ = ch;
    *s = 0;
  }

  void read(char &c) { for (c = gc(); blank(c); c = gc()); }

  void push(const char &c) {
#if DEBUG  // 调试，可显示字符
    putchar(c);
#else
    if (pp - pbuf == MAXSIZE) fwrite(pbuf, 1, MAXSIZE, stdout), pp = pbuf;
    *pp++ = c;
#endif
  }

  template <class T>
  void write(T x) {
    if (x < 0) x = -x, push('-');  // 负数输出
    static T sta[35];
    T top = 0;
    do {
      sta[top++] = x % 10, x /= 10;
    } while (x);
    while (top) push(sta[--top] + '0');
  }

  template <class T>
  void write(T x, char lastChar) {
    write(x), push(lastChar);
  }
} io;

lint qpow(lint a, lint b) {
  lint ans = 1;
  for (; b; b >>= 1) {
    if (b & 1) (ans *= a) %= p;
    (a *= a) %= p;
  }
  return ans;
}

void get_inv() {
  s[0] = 1;
  for (int i = 1; i <= n; ++i)
    s[i] = s[i - 1] * a[i] % p;
  t[n] = qpow(s[n], p - 2);
  for (int i = n - 1; ~i; --i)
    t[i] = a[i + 1] * t[i + 1] % p;
  for (int i = 1; i <= n; ++i)
    inv[i] = s[i - 1] * t[i];
}

signed main() {
  io.read(n); io.read(p); io.read(k);
  for (int i = 1; i <= n; ++i) io.read(a[i]);
  get_inv();

  for (int i = 1; i <= n; ++i) {
    (ans += qpow(k, i) * inv[i] % p) %= p;
  }
  io.write(ans, '\n');
  return 0;
}