// 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)
{
	char c, *p = s;
	do c = gc(), *s-- = c & 0xf;
	while (c > ' ');
	return p - ++s;
}

#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];
}


//// 10進数を2進数に変換
int dec2bin(char *bin, int dlen, char *dec)
{
	int  i;
	int  d, r;
	int  blen;
	int  non_zero;

	blen = 0;
	do {
		r = 0;
		non_zero = 0;
		for (i = dlen - 1; i >= 0; i--) {
			d = dec[i];
			dec[i] = d >> 1;
			if (r > 0) dec[i] += 5;
			if (dec[i] > 0) non_zero = 1;
			r = d & 1;
		}
		bin[blen++] = r;
	} while (non_zero);
	return blen;
}

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

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

char d[210]; int wd;  // 10進数
char b[600]; int wb;  // 2進数

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

	n = (int)in();
	ans = 1;
	while (n--) {
		c = fib(in() + 2);
		wd = ins(d+205);
		wb = dec2bin(b, wd, 206+d-wd);
		ans = (ans * big_mod(c, wb, b)) % MOD;
	}
	printf("%d\n", (int)ans);
	return 0;
}