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;
long n, m, p, a, b, c;
long long d[100117], arr[100117], ps[100117], dp[117][100117];
vector<pair<long long, long long> > q;
bool conv(pair<long long, long long> l1, pair<long long, long long> l2,
          pair<long long, long long> l3) {
  return (l2.first - l1.first) * (l1.second - l3.second) <=
         (l1.second - l2.second) * (l3.first - l1.first);
}
void add(long long k, long long b) {
  pair<long long, long long> x = make_pair(k, b);
  while (q.size() > 1 && conv(q[q.size() - 2], q[q.size() - 1], x))
    q.pop_back();
  q.push_back(x);
}
long long get(long long x) {
  if (c >= q.size()) c = q.size() - 1;
  while (c < q.size() - 1 && (q[c].first * x + q[c].second) >=
                                 (q[c + 1].first * x + q[c + 1].second))
    c++;
  return (q[c].first * x + q[c].second);
}
int main() {
  scanf("%ld %ld", &n, &m);
  scanf("%ld", &p);
  ;
  d[0] = 0;
  for (long i = 0; i < n - 1; ++i) {
    scanf("%I64d", &d[i + 1]);
    d[i + 1] += d[i];
  }
  for (long i = 0; i < m; ++i) {
    scanf("%ld %ld", &a, &b);
    arr[i] = b - d[a - 1];
  }
  sort(arr, arr + m);
  ps[0] = 0;
  for (long i = 0; i < m; ++i) ps[i + 1] = arr[i] + ps[i];
  for (long i = 0; i < p; ++i) dp[i][0] = 0;
  for (long j = 1; j < m + 1; ++j)
    dp[0][j] = dp[0][j - 1] + (j - 1) * (arr[j - 1] - arr[j - 2]);
  for (long i = 1; i < p; ++i) {
    q.clear();
    add(0, 0);
    c = 0;
    for (long j = 1; j < m + 1; ++j) {
      dp[i][j] = j * arr[j - 1] - ps[j] + get(arr[j - 1]);
      add(-j, dp[i - 1][j] + ps[j]);
    }
  }
  printf("%I64d", dp[p - 1][m]);
  return 0;
}
