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 long dp[110][100010];
long long a[100010];
long long s[100010];
long long q[100010];
int d[100010];
int main() {
  int n, m, p;
  scanf("%d%d%d", &n, &m, &p);
  d[1] = 0;
  for (int i = 2; i <= n; i++) {
    scanf("%d", &d[i]);
    d[i] += d[i - 1];
  }
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    y -= d[x];
    a[i] = y;
  }
  sort(a + 1, a + m + 1);
  s[0] = 0;
  for (int i = 1; i <= m; i++) s[i] = s[i - 1] + a[i];
  memset(dp, 0, sizeof(dp));
  dp[1][0] = 0;
  for (int i = 1; i <= m; i++) dp[1][i] = i * a[i] - s[i];
  for (int i = 2; i <= p; i++) {
    q[1] = 0;
    int head = 1, tail = 1;
    for (int j = 1; j <= m; j++) {
      while ((head < tail) && ((dp[i - 1][q[head + 1]] + s[q[head + 1]]) -
                                   (dp[i - 1][q[head]] + s[q[head]]) <=
                               (q[head + 1] - q[head]) * a[j]))
        head++;
      dp[i][j] =
          dp[i - 1][q[head]] + (j - q[head]) * a[j] - (s[j] - s[q[head]]);
      q[++tail] = j;
      while (head < tail - 1 &&
             ((dp[i - 1][q[tail]] + s[q[tail]]) -
              (dp[i - 1][q[tail - 1]] + s[q[tail - 1]])) *
                     (q[tail - 1] - q[tail - 2]) <=
                 ((dp[i - 1][q[tail - 1]] + s[q[tail - 1]]) -
                  (dp[i - 1][q[tail - 2]] + s[q[tail - 2]])) *
                     (q[tail] - q[tail - 1])) {
        tail--;
        q[tail] = q[tail + 1];
      }
    }
  }
  printf("%I64d\n", dp[p][m]);
  return 0;
}
