Zxr960115 is owner of a large farm. He feeds m cute cats and employs p feeders. There's a straight road across the farm and n hills along the road, numbered from 1 to n from left to right. The distance between hill i and (i - 1) is di meters. The feeders live in hill 1.

One day, the cats went out to play. Cat i went on a trip to hill hi, finished its trip at time ti, and then waited at hill hi for a feeder. The feeders must take all the cats. Each feeder goes straightly from hill 1 to n without waiting at a hill and takes all the waiting cats at each hill away. Feeders walk at a speed of 1 meter per unit time and are strong enough to take as many cats as they want.

For example, suppose we have two hills (d2 = 1) and one cat that finished its trip at time 3 at hill 2 (h1 = 2). Then if the feeder leaves hill 1 at time 2 or at time 3, he can take this cat, but if he leaves hill 1 at time 1 he can't take it. If the feeder leaves hill 1 at time 2, the cat waits him for 0 time units, if the feeder leaves hill 1 at time 3, the cat waits him for 1 time units.

Your task is to schedule the time leaving from hill 1 for each feeder so that the sum of the waiting time of all cats is minimized.

Input

The first line of the input contains three integers n, m, p (2 ≤ n ≤ 105, 1 ≤ m ≤ 105, 1 ≤ p ≤ 100).

The second line contains n - 1 positive integers d2, d3, ..., dn (1 ≤ di < 104).

Each of the next m lines contains two integers hi and ti (1 ≤ hi ≤ n, 0 ≤ ti ≤ 109).

Output

Output an integer, the minimum sum of waiting time of all cats.

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

4 6 2
1 3 5
1 0
2 1
4 9
1 10
2 10
3 12


Output

3
#include <bits/stdc++.h>
using namespace std;
template <typename T, bool MIN>
struct ConvexHullTrickAddMonotone {
  using Line = pair<T, T>;
  deque<Line> deq;
  inline bool check(const Line &a, const Line &b, const Line &c) {
    return (b.first - a.first) * (c.second - b.second) >=
           (b.second - a.second) * (c.first - b.first);
  }
  inline T f(const Line &a, const T x) { return a.first * x + a.second; }
  void add_line(T a, T b) {
    if (!MIN) a = -a, b = -b;
    Line line(a, b);
    if (deq.empty()) return (void)deq.emplace_back(line);
    if (deq.front().first <= a) {
      if (deq.front().first == a) {
        if (deq.front().second <= b) return;
        deq.pop_front();
      }
      while (deq.size() >= 2 && check(line, deq.front(), deq[1]))
        deq.pop_front();
      deq.emplace_front(line);
    } else {
      if (deq.back().first == a) {
        if (deq.back().second <= b) return;
        deq.pop_back();
      }
      while (deq.size() >= 2 && check(deq[deq.size() - 2], deq.back(), line))
        deq.pop_back();
      deq.emplace_back(line);
    }
  }
  T query(T x) {
    int l = -1, r = deq.size() - 1;
    while (l + 1 < r) {
      int m = (l + r) >> 1;
      (f(deq[m], x) >= f(deq[m + 1], x) ? l : r) = m;
    }
    return MIN ? f(deq[r], x) : -f(deq[r], x);
  }
  T query_monotone_inc(T x) {
    while (deq.size() >= 2 && f(deq.front(), x) >= f(deq[1], x))
      deq.pop_front();
    return MIN ? f(deq.front(), x) : -f(deq.front(), x);
  }
  T query_monotone_dec(T x) {
    while (deq.size() >= 2 && f(deq.back(), x) >= f(deq[deq.size() - 2], x))
      deq.pop_back();
    return MIN ? f(deq.back(), x) : -f(deq.back(), x);
  }
};
const long long INF = 1LL << 60;
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
int main() {
  int n, m;
  cin >> n >> m;
  int p;
  cin >> p;
  vector<int> d(n);
  for (int i = 0; i < n - 1; i++) {
    int x;
    cin >> x;
    d[i + 1] = d[i] + x;
  }
  vector<long long> x(m + 1);
  for (int i = 0; i < m; i++) {
    int h, t;
    cin >> h >> t;
    h--;
    x[i + 1] = t - d[h];
  }
  sort(begin(x) + 1, end(x));
  vector<vector<long long>> dp(p + 1, vector<long long>(m + 1, INF));
  for (int i = 0; i <= p; i++) dp[i][0] = 0;
  for (int i = 0; i < p; i++) {
    ConvexHullTrickAddMonotone<long long, true> cht;
    for (int j = 1; j <= m; j++) {
      cht.add_line(-(j - 1), dp[i][j - 1]);
      dp[i + 1][j] = cht.query_monotone_inc(x[j]) + x[j] * j;
    }
  }
  long long ans = dp[p][m];
  for (int i = 0; i <= m; i++) ans -= x[i];
  cout << ans << endl;
  return 0;
}
