cf-264C

/*   dp[c] := the maximal value of a sequence whose last ball's color is c

For each ball i, we want to update the array. Let the i-th ball's color be col[i], the i-th ball's value be val[i], and the maximal value of dp array other than dp[col[i]] be otherMAX. We can update the value of dp[col[i]] to dp[col[i]] + val[i] × a or otherMAX + val[i] × b. Here, we only need to know dp[col[i]] and otherMAX. If we remember the biggest two values of dp array in that time and their indexes in the array, otherMAX can be calculated using the biggest two values, which always include maximal values of dp array other than any particular color.

Since the values of dp array don't decrease, we can update the biggest two values in O(1). Finally, the answer for the query is the maximal value of dp array.

The complexity of the described algorithm is O(QN).*/


ll dp[MAX],v[MAX];ll a,b;int len;int pk[MAX],arr[MAX],col[MAX],vis[MAX][2],last[MAX];


ll giveres(int n)
{
    ///cout<<pos<<" "<<pk[pos]<<endl;

    for(int i = 1;i<MAX;i++)dp[i] = -3e17;
    pll x1 = pll(-3e17,-1),x2 = pll(-3e17,-1);
    for(int i = 1;i<=n;i++)
    {
        if(dp[col[i]]==-3e17)
        {
            dp[col[i]] = b * v[i];
        }
        else
        {
            dp[col[i]] = max(dp[col[i]],dp[col[i]] + a*v[i]);
            dp[col[i]] = max(dp[col[i]],b*v[i]);
        }
        if(x1.ss!=col[i]&&x1.ff!=-3e17)
        {
            if(x1.ff + b*v[i]>dp[col[i]])
            {
                dp[col[i]] = x1.ff + b*v[i];
            }
        }
        if(x2.ss!=col[i]&&x2.ff!=-3e17)
        {
            if(x2.ff + b*v[i]>dp[col[i]])
            {
                dp[col[i]] = x2.ff + b*v[i];
            }
        }
        if(dp[col[i]]>x1.ff)
        {
            if(x1.ss!=col[i])
            {
                swap(x1,x2);
            }
            x1.ff = dp[col[i]];
            x1.ss = col[i];
        }
        else if(dp[col[i]]>x2.ff&&x1.ss!=col[i])
        {
            x2.ff = dp[col[i]];
            x2.ss = col[i];
        }

        ///cout<<x1.ff<<" "<<x1.ss<<" "<<x2.ff<<" "<<x2.ss<<" "<<a*v[i]<<" "<<b*v[i]<<" "<<dp[col[i]]<<" "<<col[i]<<endl;

    }

    return *max_element(dp,dp+MAX);


}

int main()
{
    ///booster()
    ///read("input.txt");
    int n,q;
    scani2(n,q);
    len = n;
    for(int i = 1;i<=n;i++)scanl(v[i]);
    for(int i = 1;i<=n;i++)
    {
        scani(col[i]);
    }

///    for(int i = 1;i<=n;i++)cout<<pk[i]<<" "<<last[i]<<" "<<i<<endl;

    while(q--)
    {
        scanf("%I64d%I64d",&a,&b);
        printf("%I64d\n",giveres(n));
    }

    return 0;
}