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, x, p, d[100005], id[100005], o[100005], a[100005], q[100005];
long long s[100005], f[100005], g[100005];
bool cmp(const int i, const int j) { return o[i] < o[j]; }
int main() {
  scanf("%d%d%d", &n, &m, &p);
  for (int i = 2; i <= n; i++) scanf("%d", d + i), d[i] += d[i - 1];
  for (int i = 1; i <= m; i++) scanf("%d%d", &x, a + i), a[i] -= d[x];
  sort(a + 1, a + m + 1);
  for (int i = 1; i <= m; i++) s[i] = s[i - 1] + a[i];
  for (int i = 1; i <= m; i++) f[i] = 1LL * a[i] * i - s[i];
  for (int j = 2; j <= p; j++) {
    int h = 0, t = 0;
    q[t++] = 0;
    for (int i = 1; i <= m; i++) {
      while (t - 2 >= h && (f[i] - (f[q[t - 1]] + 1LL * a[i] * (i - q[t - 1]) -
                                    s[i] + s[q[t - 1]])) *
                                   (q[t - 1] - q[t - 2]) <=
                               ((f[q[t - 1]] + 1LL * a[i] * (i - q[t - 1]) -
                                 s[i] + s[q[t - 1]]) -
                                (f[q[t - 2]] + 1LL * a[i] * (i - q[t - 2]) -
                                 s[i] + s[q[t - 2]])) *
                                   (i - q[t - 1]))
        t--;
      q[t++] = i;
      while (h + 1 <= t - 1 &&
             f[q[h]] + 1LL * a[i] * (i - q[h]) - s[i] + s[q[h]] >=
                 f[q[h + 1]] + 1LL * a[i] * (i - q[h + 1]) - s[i] + s[q[h + 1]])
        h++;
      g[i] = min(f[i], f[q[h]] + 1LL * a[i] * (i - q[h]) - s[i] + s[q[h]]);
    }
    for (int i = 1; i <= m; i++) f[i] = g[i];
  }
  printf("%I64d\n", f[m]);
  return 0;
}
