Problem: There are n balls. They are arranged in a row. Each ball has a color (for convenience an integer) and an integer value. The color of the i-th ball is ci and the value of the i-th ball is vi.

Squirrel Liss chooses some balls and makes a new sequence without changing the relative order of the balls. She wants to maximize the value of this sequence.

The value of the sequence is defined as the sum of following values for each ball (where a and b are given constants):

  * If the ball is not in the beginning of the sequence and the color of the ball is same as previous ball's color, add (the value of the ball)  ×  a. 
  * Otherwise, add (the value of the ball)  ×  b. 



You are given q queries. Each query contains two integers ai and bi. For each query find the maximal value of the sequence she can make when a = ai and b = bi.

Note that the new sequence can be empty, and the value of an empty sequence is defined as zero.

Input

The first line contains two integers n and q (1 ≤ n ≤ 105; 1 ≤ q ≤ 500). The second line contains n integers: v1, v2, ..., vn (|vi| ≤ 105). The third line contains n integers: c1, c2, ..., cn (1 ≤ ci ≤ n).

The following q lines contain the values of the constants a and b for queries. The i-th of these lines contains two integers ai and bi (|ai|, |bi| ≤ 105).

In each line integers are separated by single spaces.

Output

For each query, output a line containing an integer — the answer to the query. The i-th line contains the answer to the i-th query in the input order.

Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

6 3
1 -2 3 4 0 -1
1 2 1 2 1 1
5 1
-2 1
1 0


Output

20
9
4


Input

4 1
-3 6 -1 2
1 2 3 1
1 -1


Output

5

Note

In the first example, to achieve the maximal value:

  * In the first query, you should select 1st, 3rd, and 4th ball. 
  * In the second query, you should select 3rd, 4th, 5th and 6th ball. 
  * In the third query, you should select 2nd and 4th ball. 



Note that there may be other ways to achieve the maximal value.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class _T>
inline _T sqr(const _T &first) {
  return first * first;
}
template <class _T>
inline string tostr(const _T &a) {
  ostringstream os("");
  os << a;
  return os.str();
}
const long double PI = 3.1415926535897932384626433832795L;
const long double EPS = 1e-8;
char TEMPORARY_CHAR;
const int INF = 2e9;
inline void fft(vector<complex<double> > &a, bool invert) {
  int n = (int)a.size();
  for (int i = 1, j = 0; i < n; ++i) {
    int bit = n >> 1;
    for (; j >= bit; bit >>= 1) j -= bit;
    j += bit;
    if (i < j) swap(a[i], a[j]);
  }
  for (int len = 2; len <= n; len <<= 1) {
    double ang = 2 * (double)PI / len * (invert ? -1 : 1);
    complex<double> wlen(cos(ang), sin(ang));
    for (int i = 0; i < n; i += len) {
      complex<double> w(1);
      for (int j = 0; j < len / 2; ++j) {
        complex<double> u = a[i + j], v = a[i + j + len / 2] * w;
        a[i + j] = u + v;
        a[i + j + len / 2] = u - v;
        w *= wlen;
      }
    }
  }
  if (invert)
    for (int i = 0; i < n; ++i) a[i] /= n;
}
inline void input(int &a) {
  a = 0;
  while (((TEMPORARY_CHAR = getchar()) > '9' || TEMPORARY_CHAR < '0') &&
         (TEMPORARY_CHAR != '-')) {
  }
  char neg = 0;
  if (TEMPORARY_CHAR == '-') {
    neg = 1;
    TEMPORARY_CHAR = getchar();
  }
  while (TEMPORARY_CHAR <= '9' && TEMPORARY_CHAR >= '0') {
    a = (a << 3) + (a << 1) + TEMPORARY_CHAR - '0';
    TEMPORARY_CHAR = getchar();
  }
  if (neg) a = -a;
}
inline void out(int a) {
  if (!a) putchar('0');
  if (a < 0) {
    putchar('-');
    a = -a;
  }
  char s[10];
  int i;
  for (i = 0; a; ++i) {
    s[i] = '0' + a % 10;
    a /= 10;
  }
  for (int j = (i)-1; j >= 0; j--) putchar(s[j]);
}
inline int nxt() {
  int(ret);
  input((ret));
  ;
  return ret;
}
struct lnum {
  vector<int> a;
  int base;
  lnum(int num = 0, int base = 1000000000) : base(base) {
    if (!num) a.resize(1);
    while (num) {
      a.push_back(num % base);
      num /= base;
    }
  }
  inline int len() const { return a.size(); }
  lnum &operator=(const lnum &l) {
    if (this != &l) {
      a = l.a;
      base = l.base;
    }
    return *this;
  }
  inline friend lnum operator+(const lnum &l, const lnum &r) {
    lnum ret(0, l.base);
    int base = l.base;
    int ln = l.len(), rn = r.len();
    int n = max(ln, rn);
    ret.a.resize(n);
    int o = 0;
    for (int i = 0; i < n; ++i) {
      int s = o;
      if (i < ln) s += l.a[i];
      if (i < rn) s += r.a[i];
      o = s >= base;
      if (o) s -= base;
      ret.a[i] = s;
    }
    if (o) ret.a.push_back(1);
    return ret;
  }
  inline friend lnum operator-(const lnum &l, const lnum &r) {
    lnum ret(0, l.base);
    int base = l.base;
    int n = l.len();
    int rn = r.len();
    ret.a.resize(n);
    int o = 0;
    for (int i = 0; i < n; ++i) {
      int s = l.a[i] - o;
      if (i < rn) s -= r.a[i];
      o = s < 0;
      if (o) s += base;
      ret.a[i] = s;
    }
    if (ret.len() > 1 && !ret.a.back()) ret.a.pop_back();
    return ret;
  }
  inline friend lnum operator*(const lnum &l, const lnum &r) {
    lnum ret(0, l.base);
    int base = l.base;
    if (l.len() * r.len() > 1000000) {
      vector<complex<double> > fa(l.a.begin(), l.a.end()),
          fb(r.a.begin(), r.a.end());
      int n = 1;
      while (n < max(l.len(), r.len())) n <<= 1;
      n <<= 1;
      fa.resize(n), fb.resize(n);
      fft(fa, false), fft(fb, false);
      for (int i = 0; i < n; ++i) fa[i] *= fb[i];
      fft(fa, true);
      ret.a.resize(n);
      for (int i = 0; i < n; ++i) ret.a[i] = int(fa[i].real() + 0.5);
      int carry = 0;
      for (int i = 0; i < n; ++i) {
        ret.a[i] += carry;
        carry = ret.a[i] / base;
        ret.a[i] %= base;
      }
    } else {
      ret.a.resize(l.len() + r.len());
      for (int i = 0; i < l.len(); ++i)
        for (int j = 0, carry = 0; j < r.len() || carry; ++j) {
          long long cur = ret.a[i + j] +
                          (long long)l.a[i] * (j < r.len() ? r.a[j] : 0) +
                          carry;
          ret.a[i + j] = cur % base;
          carry = cur / base;
        }
    }
    while (ret.len() > 1 && !ret.a.back()) ret.a.pop_back();
    return ret;
  }
  inline friend lnum operator/(const lnum &l, const int &r) {
    lnum ret(0, l.base);
    ret.a.resize(l.len());
    int carry = 0;
    for (int i = l.len() - 1; i >= 0; --i) {
      long long cur = l.a[i] + (long long)carry * l.base;
      ret.a[i] = cur / r;
      carry = cur % r;
    }
    while (ret.len() > 1 && ret.a.back() == 0) ret.a.pop_back();
    return ret;
  }
  inline friend bool operator<(const lnum &l, const lnum &r) {
    if (l.len() < r.len()) return true;
    if (l.len() > r.len()) return false;
    int n = l.len();
    for (int i = n - 1; i; --i) {
      if (l.a[i] < r.a[i]) return true;
      if (l.a[i] > r.a[i]) return false;
    }
    return false;
  }
  inline friend bool operator>(const lnum &l, const lnum &r) { return r > l; }
  inline friend bool operator==(const lnum &l, const lnum &r) {
    if (l.len() != r.len()) return false;
    int n = l.len();
    for (int i = n - 1; i; --i) {
      if (l.a[i] != r.a[i]) return false;
    }
    return true;
  }
  inline friend bool operator!=(const lnum &l, const lnum &r) {
    return !(l == r);
  }
  inline void print() {
    if (base == 1000000000) {
      printf("%d", a.back());
      for (int i = a.size() - 2; i >= 0; --i) printf("%09d", a[i]);
    } else {
      for (int i = a.size() - 1; i >= 0; --i) printf("%d", a[i]);
    }
  }
};
int gcd(int a, int b) {
  if (a > b) swap(a, b);
  if (!a) return a;
  if (a == b) return a;
  if (a == 1) return 1;
  if (!(a & 1)) {
    if (b & 1)
      return gcd(a >> 1, b);
    else
      return gcd(a >> 1, b >> 1) << 1;
  } else {
    if (b & 1)
      return gcd((b - a) >> 1, a);
    else
      return gcd(a, b >> 1);
  }
}
int main() {
  int(n);
  input((n));
  ;
  int(q);
  input((q));
  ;
  long long v[n];
  int c[n];
  for (int i = 0; i < (int)(n); i++) v[i] = nxt();
  for (int i = 0; i < (int)(n); i++) c[i] = nxt() - 1;
  while (q--) {
    long long a, b;
    a = nxt();
    b = nxt();
    long long dp[n];
    char used[n];
    memset((used), 0, sizeof(used));
    for (int i = 0; i < (int)(n); i++) dp[i] = -1e18;
    vector<pair<long long, int> > best;
    for (int i = 0; i < (int)(n); i++) {
      long long best1 = dp[c[i]], best2 = 0;
      for (int j = 0; j < (int)(best.size()); j++) {
        if (best[j].second != c[i])
          ((best2) = (best[j].first) > (best2) ? (best[j].first) : (best2));
      }
      long long res;
      if (used[c[i]])
        res = max(best1, max(best1 + a * v[i], best2 + b * v[i]));
      else
        res = best2 + b * v[i];
      used[c[i]] = 1;
      ((dp[c[i]]) = (res) > (dp[c[i]]) ? (res) : (dp[c[i]]));
      bool flag = true;
      for (int j = 0; j < (int)(best.size()); j++) {
        if (best[j].second == c[i]) {
          ((best[j].first) = (res) > (best[j].first) ? (res) : (best[j].first));
          flag = false;
        }
      }
      if (flag) {
        best.push_back(make_pair(res, c[i]));
        sort((best).begin(), (best).end());
        reverse((best).begin(), (best).end());
        best.resize(min(2, (int)best.size()));
      }
    }
    long long ret = 0;
    for (int i = 0; i < (int)(n); i++)
      ((ret) = (dp[i]) > (ret) ? (dp[i]) : (ret));
    cout << ret << endl;
  }
  return 0;
}