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 a[100001], i, j, m, n, p, k, t[100001];
long long cost[100001], f[101][100001], s[100001], que[100001];
long long up(int x, int y) { return f[i - 1][x] + s[x] - (f[i - 1][y] + s[y]); }
long long down(int x, int y) { return x - y; }
int main() {
  scanf("%d%d%d", &n, &m, &p);
  for (i = 2; i <= n; i++) scanf("%d", &a[i]), a[i] += a[i - 1];
  for (i = 1; i <= m; i++) {
    scanf("%d%d", &k, &t[i]);
    t[i] -= a[k];
  }
  sort(t + 1, t + m + 1);
  for (i = 1; i <= m; i++) s[i] = s[i - 1] + t[i];
  memset(f, 63, sizeof(f));
  f[0][0] = 0;
  if (p >= m) {
    printf("0\n");
    return 0;
  }
  for (i = 1; i <= p; i++) {
    int l = 1, r = 1;
    que[1] = 0;
    for (j = 1; j <= m; j++) {
      while (l < r &&
             up(que[l + 1], que[l]) <= 1ll * t[j] * down(que[l + 1], que[l]))
        l++;
      f[i][j] = f[i - 1][que[l]] + t[j] * (j - que[l]) + s[que[l]] - s[j];
      while (l < r && up(que[r], que[r - 1]) * down(j, que[r]) >
                          up(j, que[r]) * down(que[r], que[r - 1]))
        r--;
      que[++r] = j;
    }
  }
  printf("%I64d\n", f[p][m]);
}
