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 dp[100001], s[100001], sum[100001], g[100001];
struct Node {
  int id;
  long long num;
} q[100001];
int head, tail;
int main() {
  memset(dp, 63, sizeof(dp));
  scanf("%d%d%d", &n, &m, &p);
  for (int i = 2; i <= n; i++) {
    long long x;
    scanf("%lld", &x);
    s[i] = s[i - 1] + x;
  }
  for (int i = 1; i <= m; i++) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    g[i] = y - s[x];
  }
  sort(g + 1, g + 1 + m);
  for (int i = 1; i <= m; i++) sum[i] = sum[i - 1] + g[i];
  dp[0] = 0;
  for (int i = 1; i <= p; i++) {
    head = tail = 1;
    q[1] = (Node){0, 0};
    for (int j = 1; j <= m; j++) {
      while (head < tail && g[j] * (q[head + 1].id - q[head].id) >=
                                (q[head + 1].num - q[head].num))
        head++;
      while (head < tail &&
             (dp[j] + sum[j] - q[tail].num) * (q[tail].id - q[tail - 1].id) <=
                 (q[tail].num - q[tail - 1].num) * (j - q[tail].id))
        tail--;
      q[++tail] = (Node){j, dp[j] + sum[j]};
      dp[j] = q[head].num - sum[j] + g[j] * (j - q[head].id);
    }
  }
  printf("%lld", dp[m]);
  return 0;
}
