/*
CodeForces 840C - On the Bench [ DP ]  |  Codeforces Round #429 (Div. 1)
题意：给出一个数组，问有多少种下标排列，使得任意两个相邻元素的乘积不是完全平方数
分析：将数组分组，使得每组中的任意两个数之积为完全平方数由唯一分解定理可知，每个质因子的幂次的奇偶性相同的两个数之积为完全平方数即按每个质因子的幂次的奇偶性分组，故这样的分组唯一然后问题归结于每组中的数不能相邻的排列有几种设 dp[i][j]表示 前i组相邻的同组的数有j对考虑把第i+1组分段后插入前i组的空隙中枚举将下一组分成k段，每段相邻枚举k段中有l段插在前面j对同组的空隙中设前i组总个数为sum, 第i+1组个数为num则得到转移方程dp[i+1][j-l+num-k] += C(num-1, k-1) * C(j, l) * C(sum+1-j, k-l) * dp[i][j]组合数什么的仔细推导下，再最后乘上每组的排列数
*/
#include <bits/stdc++.h>
using namespace std;
#define LL long long
const int MOD = 1e9+7;
const int N = 305;
LL C[N][N], F[N];
void init() {C[0][0] = 1;for (int i = 1; i < N; i++) {C[i][0] = C[i][i] = 1;for (int j = 1; j < i; j++)C[i][j] = (C[i-1][j] + C[i-1][j-1]) % MOD;}F[0] = 1;for (int i = 1; i < N; i++) F[i] = i * F[i-1] % MOD;
}
bool check(LL a, LL b)
{LL l = 1, r = 1e10, mid;while (l <= r){mid = (l+r) >> 1;if (mid*mid <= a*b) l = mid+1;else r = mid-1;}return r*r == a*b;
}
LL dp[N][N], ans;
int n, a[N], id[N], num[N], cnt;
void solve()
{dp[0][0] = 1;int sum = 0;for (int i = 1; i <= cnt; i++)//第i组{for (int j = 0; j <= sum; j++)//j处平方for (int k = 1; k <= num[i]; k++)//num[i]分成k段for (int l = 0; l <= j && l <= k; l++)//j 中 l 段{LL tmp = dp[i-1][j];tmp = tmp * C[num[i]-1][k-1] % MOD;tmp = tmp * C[j][l] % MOD;tmp = tmp * C[sum+1-j][k-l] % MOD;dp[i][j-l+num[i]-k] += tmp;dp[i][j-l+num[i]-k] %= MOD;}sum += num[i];}ans = dp[cnt][0];for (int i = 1; i <= cnt; i++) ans = ans * F[num[i]] % MOD;
}
int main()
{init();cnt = 0;scanf("%d", &n);for (int i = 1; i <= n; i++){scanf("%d", &a[i]);bool flag = 0;for (int j = 1; j < i; j++){if (check(a[i], a[j])){num[id[j]]++;id[i] = id[j];flag = 1; break;}}if (!flag){id[i] = ++cnt;num[cnt] = 1;}}solve();printf("%lld\n", ans);
}