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;
const long long INF = 1e18;
struct line {
  long long k, m;
  line(long long _k, long long _m) {
    k = _k;
    m = _m;
  }
  long double inters(line a) { return (a.m * 1.0 - m) / (k - a.k); }
  long double v(long double x) { return k * x + m; }
};
struct CHT {
  vector<long double> p;
  vector<line> l;
  void add(line _l) {
    if (l.empty()) {
      l.push_back(_l);
      return;
    }
    bool f = 1;
    while (f && !p.empty()) {
      long double x = p.back();
      if (_l.v(x) <= l.back().v(x))
        p.pop_back(), l.pop_back();
      else
        f = 0;
    }
    p.push_back(_l.inters(l.back()));
    l.push_back(_l);
  };
  long long mv(long long x) {
    int i = lower_bound(p.begin(), p.end(), 1.0 * x) - p.begin();
    return l[i].k * x + l[i].m;
  }
  void clear() { l.clear(), p.clear(); }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, p;
  cin >> n >> m >> p;
  vector<long long> d(n);
  d[0] = 0;
  for (int i = 1; i < n; ++i) {
    cin >> d[i];
    d[i] += d[i - 1];
  }
  vector<long long> s(m + 1);
  vector<pair<long long, long long> > _q(m);
  vector<long long> q;
  q.push_back(-INF);
  for (int i = 0; i < m; ++i) {
    cin >> _q[i].first >> _q[i].second;
    _q[i].first--;
  }
  for (int i = 0; i < m; ++i) q.push_back(_q[i].second - d[_q[i].first]);
  sort(q.begin(), q.end());
  for (int i = 1; i <= m; ++i) s[i] = s[i - 1] + q[i];
  vector<vector<long long> > dp(p + 1, vector<long long>(m + 1, INF));
  dp[0][0] = 0;
  CHT t;
  for (int i = 1; i <= p; ++i) {
    dp[i][0] = 0;
    for (int j = 1; j <= m; ++j) {
      t.add(line(-(j - 1), s[j - 1] + dp[i - 1][j - 1]));
      dp[i][j] = t.mv(q[j]) + j * q[j] - s[j];
    }
    t.clear();
  }
  cout << dp[p][m];
  return 0;
}
