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 int INF = 1000 * 1000 * 1000 + 7;
const double EPS = 1e-9;
const double PI = acos(-1.0);
template <class T>
int chmin(T &t, T f) {
  return (t > f) ? (t = f, 1) : 0;
}
template <class T>
int chmax(T &t, T f) {
  return (t < f) ? (t = f, 1) : 0;
}
inline int getint() {
  int a;
  return scanf("%d", &a) ? a : (fprintf(stderr, "trying to read\n"), -1);
}
inline double getdouble() {
  double a;
  return scanf("%lf", &a) ? a : (fprintf(stderr, "trying to read\n"), -1);
}
pair<int, long long> operator-(pair<int, long long> a, pair<int, long long> b) {
  return pair<int, long long>(a.first - b.first, a.second - b.second);
}
long double operator^(pair<int, long long> a, pair<int, long long> b) {
  return (long double)a.first * b.second - (long double)a.second * b.first;
}
int sign(long double first) { return (first > 0) - (first < 0); }
int ccw(pair<int, long long> a, pair<int, long long> b,
        pair<int, long long> c) {
  return sign((a - c) ^ (b - c));
}
const int HILLS = 1 << 17, FEEDERS = 1 << 7;
int distHill[HILLS];
long long dp[FEEDERS][HILLS];
long long sreq[HILLS];
void myCode() {
  int hills = getint(), cats = getint(), feeders = getint();
  distHill[0] = 0;
  for (int i = (int)1; i < (int)hills; i++)
    distHill[i] = distHill[i - 1] + getint();
  vector<long long> requests;
  for (int i = (int)0; i < (int)cats; i++) {
    int h = getint() - 1, t = getint();
    requests.push_back(t - distHill[h]);
  }
  sort((requests).begin(), (requests).end());
  sreq[0] = requests[0];
  for (int i = (int)1; i < (int)((int)(requests).size()); i++)
    sreq[i] = sreq[i - 1] + requests[i];
  for (int i = (int)0; i < (int)feeders + 1; i++)
    for (int j = (int)0; j < (int)hills + 1; j++) dp[i][j] = 1LL << 63;
  for (int j = (int)0; j < (int)((int)(requests).size()); j++)
    dp[1][j] = (j + 1) * requests[j] - sreq[j];
  for (int i = (int)2; i < (int)feeders + 1; i++) {
    deque<pair<int, long long> > q;
    for (int j = (int)0; j < (int)((int)(requests).size()); j++) {
      dp[i][j] = dp[i - 1][j];
      while (((int)(q).size()) >= 2 &&
             (q[1].second - q[0].second) <=
                 requests[j] * (q[1].first - q[0].first))
        q.pop_front();
      if (!q.empty()) {
        int k = q.front().first;
        chmin(dp[i][j], dp[i - 1][k] - k * requests[j] + sreq[k] - sreq[j] +
                            j * requests[j]);
      }
      while (((int)(q).size()) >= 2 &&
             ccw(q[((int)(q).size()) - 2], q.back(),
                 make_pair(j, dp[i - 1][j] + sreq[j])) <= 0)
        q.pop_back();
      q.push_back(make_pair(j, dp[i - 1][j] + sreq[j]));
    }
  }
  cout << dp[feeders][cats - 1] << endl;
}
int main() {
  myCode();
  return 0;
}
