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 n, m, p;
long long int a[100002];
long long int cost[2][100002];
long long int sum[100002];
int d[100002];
int q[100002];
long long int f(int prev, int x) { return cost[prev][q[x]] + sum[q[x]]; }
long long int df(int prev, int x, int y) { return f(prev, x) - f(prev, y); }
bool mono(int j, int base, int prev) {
  return (cost[prev][q[base + 1]] + sum[q[base + 1]]) -
             (cost[prev][q[base]] + sum[q[base]]) <
         (q[base + 1] - q[base]) * a[j];
}
int main() {
  cin >> n >> m >> p;
  d[1] = 0;
  for (int i = 2; i <= n; d[i] += d[i - 1], i++) cin >> d[i];
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    a[i] = y - d[x];
  }
  sort(a + 1, a + m + 1);
  sum[0] = 0;
  for (int i = 1; i <= m; i++) sum[i] = sum[i - 1] + a[i];
  int now = 0, prev = 1;
  for (int i = 1; i <= m; i++) cost[now][i] = i * a[i] - sum[i];
  for (int i = 2; i <= p; i++) {
    q[1] = 0;
    int base = 1, top = 1;
    swap(now, prev);
    for (int j = 1; j <= m; j++) {
      while (base < top && mono(j, base, prev)) base++;
      cost[now][j] =
          cost[prev][q[base]] + (j - q[base]) * a[j] - (sum[j] - sum[q[base]]);
      q[++top] = j;
      while (base < top - 1 &&
             df(prev, top, top - 1) * (q[top - 1] - q[top - 2]) <
                 df(prev, top - 1, top - 2) * (q[top] - q[top - 1])) {
        top--;
        q[top] = q[top + 1];
      }
    }
  }
  cout << cost[now][m] << endl;
}
