// yukicoder: No.147 試験監督（2）
// 2019.4.11 bal4u

#include <stdio.h>
#include <stdlib.h>

//// 高速入力
#if 1
#define gc() getchar_unlocked()
#else
#define gc() getchar()
#endif
long long in()
{
	int c = gc();
	long long n = 0;
	do n = 10 * n + (c & 0xf), c = gc(); while (c >= '0');
	return n;
}

int ins(char *s)  // 文字列の入力　スペース以下の文字で入力終了
{
	char *p = s;
	do *s = gc();
	while (*s++ > ' ');
	*--s = 0;
	return s - p;
}


#define MOD 1000000007

//// fibonacci数の高速計算
void multiply(int F[2][2], int M[2][2])
{
	int x = ((long long)F[0][0] * M[0][0] + (long long)F[0][1] * M[1][0]) % MOD;
	int y = ((long long)F[0][0] * M[0][1] + (long long)F[0][1] * M[1][1]) % MOD;
	int z = ((long long)F[1][0] * M[0][0] + (long long)F[1][1] * M[1][0]) % MOD;
	int w = ((long long)F[1][0] * M[0][1] + (long long)F[1][1] * M[1][1]) % MOD;
	F[0][0] = x;
	F[0][1] = y;
	F[1][0] = z;
	F[1][1] = w;
}

void power(int F[2][2], long long n)
{
	int M[2][2] = { {1,1},{1,0} };
	if (n == 0 || n == 1) return;
	power(F, n >> 1);
	multiply(F, F);
	if (n & 1) multiply(F, M);
}

int fib(long long n)
{
	int F[2][2] = { {1,1},{1,0} };
	if (n == 0)	return 0;
	power(F, n - 1);
	return F[0][0];
}


//// 多倍長整数
#define N       1000000000
int num[50];
int qq[50], rr[2];

void mpStr2Num(int *num, int len, char *str)
{
	char *ss = str + len;
	int  k, x, *nn;

	x = 0, k = 1, nn = num;
	do {
		x += (*--ss - '0') * k;
		k *= 10;
		if (k == N || ss == str) *++nn = x, x = 0, k = 1;
	} while (ss != str);
	*num = nn - num;
}

void mpDiv(int *q, int *r, int *za, int zb)
{
	int  i;
	int  *aa, *qq;
	int  ca;
	long long x;

	*r = 0, *q = *za;
	for (ca = 0, aa = za + *za, qq = q + *q, i = *za; i > 0; i--) {
		x = (long long)N * ca + *aa--;
		*qq-- = (int)(x / zb), ca = (int)(x % zb);
	}
	if (*(q + *q) == 0) (*q)--;
	if (ca > 0) *r++ = 1, *r = ca;
	else *r = 0;
}

// calc b^p (mod m)
int big_mod(int b, int p)
{
	long long d, s = 1;

	if (b == 0) return 0;
	d = b % MOD;
	while (p) {
		if (p & 1) s = (s * d) % MOD;
		p >>= 1;
		d = (d * d) % MOD;
	}
	return (int)s;
}

char d[220];

int main()
{
	int n, c, w;
	long long ans;

	n = (int)in();
	ans = 1;
	while (n--) {
		c = fib(in() + 2);
		w = ins(d);
		mpStr2Num(num, w, d);
		mpDiv(qq, rr, num, MOD-1);
		ans = (ans * big_mod(c, rr[1])) % MOD;
	}
	printf("%d\n", (int)ans);
	return 0;
}