直接枚举左端点，先找第一个 $gcd$ 为左端点的区间，再找下一个使当前 $gcd$ 不同的区间，但是Wa了，求助

#include<bits/stdc++.h>
using namespace std;
inline long long read()
{
	char c (getchar());
	long long x (0);
	while (!isdigit(c)) c = getchar();
	while (isdigit(c)) x = (x << 1) + (x << 3) + (c & 15) , c = getchar();
	return x;
}
long long fa[100005][20];
int main()
{
	int T (read());
	while (T --)
	{
		const int n (read());
		long long ans (0);
		for (int i (1) ; i <= n ; ++ i)
		{
			fa[i][0] = read();
			for (int j (1) ; j <= 19 ; ++ j)
				fa[i][j] = 0;
		}
		for (int i (n - 1) ; i ; -- i)
			for (int j (0) ; i + (1 << j + 1) - 1 <= n ; ++ j)
				fa[i][j + 1] = __gcd(fa[i][j] , fa[i + (1 << j)][j]);
		for (int l (1) ; l <= n ; ++ l)
		{
			long long now (fa[l][0]);
			int R (l);
			for (int i (19) ; i >= 0 ; -- i)
				if (fa[R + 1][i] && fa[R + 1][i] % now == 0)
					R += 1 << i;
			ans = max(ans , (R - l + 1) * now);
			for (int r (R + 1) ; r <= n ; r = R + 1)
			{
				R = r;
				now = __gcd(now , fa[R][0]);
				for (int i (19) ; i >= 0 ; -- i)
					if (fa[R + 1][i] && fa[R + 1][i] % now == 0)
						R += 1 << i;
				ans = max (ans , (R - l + 1) * now);
			}
		}
		printf("%lld\n" , ans);
	}
	return 0;
}