#include <stdio.h>
#include <string.h>
 
int x[1000002]; // 原数列
int w[1000002]; // w[i]记录长度为i的后缀数列里面的unique元素的个数
long long dp[1000002];
int dp2[1000002]; // dp[i]记录有多少个元素的前一个相同元素的下标与该元素下标的差是i
 
int main()
{
    int n,i,cc;
	while(scanf("%d",&n)&&n) // 输入数列长度
	{
		memset(dp,0,sizeof(dp)); // 此时dp[i]记录i这个元素出现的最大下标，初始化为0
		memset(dp2,0,sizeof(dp2));
		for(i=1;i<=n;i++)
		{
			scanf("%d",x+i); // 读取原数列第i个数
			dp2[i-dp[x[i]]]++; // 获取x[i]前一个相同元素的下标、做差、存进dp2
			dp[x[i]] = i; // 更新x[i]这个元素的最大下标，存进dp
		}
		cc = 0; // 记录后缀数组中unique元素的个数，初始化为0
		memset(dp,0,sizeof(dp)); // 重用dp数组，这里dp[i]记录的是i这个元素是否出现后缀数列中
		for(i=n;i>=1;i--) // 从n枚举到1
		{
			if(dp[x[i]]==0) cc++; // 如果x[i]未出现在后缀数组中，则增加cc
			dp[x[i]]=1; // 记录已出现在后缀数组中，存进dp
			w[n-i+1] = cc; // 长度为n-i+1的后缀中unique元素的个数，存进w
		}
		dp[1] = n; // 重用dp数组，这里dp[i]表示询问为i的时候的答案，当询问是1的时候，答案必然是n
		cc = n; // 重用cc，记录当前增量
		for(i=2;i<=n;i++) // 从2到n枚举
		{
			// 询问为i的结果可以由询问为i-1的结果递推
			dp[i] = dp[i-1] - w[i-1]; // 当询问为i-1的时候，最后一个区间无法添加新的数字，
									  // 所以结果里面要减去长度为i-1的后缀数列里面的unique元素的个数
			cc -= dp2[i-1]; // 和前一个相同元素下标差为i-1的元素从这里开始不再存在增量
			dp[i] += cc; // 每个区间都多增加了一个元素，一共有cc个区间会增加一个新的unique元素
		}
		scanf("%d",&cc); // 以下为查询
		while(cc--)
		{
			scanf("%d",&i);
			printf("%I64d\n",dp[i]);
		}
	}
 
}

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