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;
int d[100001];
long long r[100001], b[100001];
int q[100001];
pair<long long, long long> ln[100001];
long long pd[2][100001];
bool pop(pair<long long, long long>& l1, pair<long long, long long>& l2,
         pair<long long, long long>& l3) {
  return (l1.first - l3.first) * (l2.second - l1.second) <=
         (l1.first - l2.first) * (l3.second - l1.second);
}
long long eval(int i, int x) { return ln[i].first + x * ln[i].second; }
int main() {
  int n, m, p;
  scanf("%d%d%d", &n, &m, &p);
  for (int i = 1; i < n; ++i) {
    scanf("%d", d + i);
    d[i] += d[i - 1];
  }
  for (int i = 1, h; i <= m; ++i) {
    cin >> h >> r[i];
    --h;
    r[i] -= d[h];
  }
  sort(r + 1, r + m + 1);
  for (int i = 1; i <= m; ++i) b[i] = r[i] + b[i - 1];
  int j = 0;
  for (int i = 1; i <= m; ++i) pd[j][i] = r[i] * i - b[i];
  j = 1;
  for (int k = 1; k < p; ++k, j ^= 1) {
    int fr = 0, bk = -1;
    for (int i = k + 1; i <= m; ++i) {
      ln[i - 1] =
          pair<long long, long long>(pd[j ^ 1][i - 1] + b[i - 1], -i + 1);
      while (fr < bk && pop(ln[q[bk - 1]], ln[q[bk]], ln[i - 1])) --bk;
      q[++bk] = i - 1;
      while (fr < bk && eval(q[fr], r[i]) >= eval(q[fr + 1], r[i])) ++fr;
      pd[j][i] = eval(q[fr], r[i]) + r[i] * i - b[i];
    }
  }
  cout << pd[j ^ 1][m] << endl;
  return 0;
}
