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;
const int MAXN = 100005, MAXP = 105;
int n, m, p, h[MAXN];
long long d[MAXN], sumd[MAXN], t[MAXN], dp[MAXN][MAXP], sum;
struct Lin {
  long long k, b;
  long long val(long long x) { return k * x + b; }
  Lin(long long _k = 0, long long _b = 0) {
    k = _k;
    b = _b;
  }
} que[MAXN];
const long long INF = (long long)1e18;
double Lin_cross_x(const Lin &A, const Lin &B) {
  return (B.b - A.b) * 1.0 / (A.k - B.k);
}
void q_push_front(int &head, int &tail, Lin x) {
  while (tail - head + 1 >= 2 &&
         Lin_cross_x(x, que[head]) > Lin_cross_x(x, que[head + 1]))
    ++head;
  que[--head] = x;
}
void q_push_back(int &head, int &tail, Lin x) {
  while (tail - head + 1 >= 2 &&
         Lin_cross_x(x, que[tail]) < Lin_cross_x(x, que[tail - 1]))
    --tail;
  que[++tail] = x;
}
long long q_get_val(int &head, int &tail, const int &x) {
  if (tail - head + 1 <= 0) return INF;
  while (tail - head + 1 >= 2 && que[head].val(x) > que[head + 1].val(x))
    ++head;
  return que[head].val(x);
}
int main() {
  scanf("%d %d %d", &n, &m, &p);
  for (int i = 2; i <= n; ++i) {
    scanf("%lld", &d[i]);
    sumd[i] = sumd[i - 1] + d[i];
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%d %lld", &h[i], &t[i]);
    t[i] -= sumd[h[i]];
    sum += t[i];
  }
  sort(t + 1, t + 1 + m);
  for (int i = 1; i <= m; ++i) {
    dp[i][1] = t[i] * i;
  }
  for (int j = 2; j <= p; ++j) {
    int head = 1, tail = 1;
    que[head] = Lin(-(j - 1), dp[j - 1][j - 1]);
    for (int i = j; i <= m; ++i) {
      dp[i][j] = q_get_val(head, tail, t[i]) + t[i] * i;
      q_push_back(head, tail, Lin(-i, dp[i][j - 1]));
    }
  }
  printf("%lld\n", dp[m][p] - sum);
  return 0;
}
