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 int n, m, g, ara[100002], dp[100002][101], sum[100002], dist[100002];
long long int cost(int i, int j) {
  return (ara[j]) * (j - i + 1) - (sum[j] - sum[i - 1]);
}
void rec(int g, int l, int r, int kl, int kr) {
  if (l > r) return;
  int mid = (l + r) / 2;
  dp[mid][g] = 1111111111111111111ll;
  int bestk = -1;
  for (int i = kl; i <= kr; i++) {
    if (i > mid) break;
    long long int now = dp[i][g - 1] + cost(i + 1, mid);
    if (now < dp[mid][g]) {
      dp[mid][g] = now;
      bestk = i;
    }
  }
  rec(g, l, mid - 1, kl, bestk);
  rec(g, mid + 1, r, bestk, kr);
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin >> n >> m >> g;
  for (int i = 2; i <= n; i++) {
    cin >> dist[i];
    dist[i] = dist[i] + dist[i - 1];
  }
  for (int i = 1; i <= m; i++) {
    long long int hi, ti;
    cin >> hi >> ti;
    ara[i] = ti - dist[hi];
  }
  sort(ara + 1, ara + m + 1);
  for (int i = 1; i <= m; i++) sum[i] = sum[i - 1] + ara[i];
  for (int i = 1; i <= m; i++) dp[i][1] = cost(1, i);
  for (int i = 2; i <= g; i++) {
    rec(i, 1, m, 1, m);
  }
  cout << dp[m][g] << endl;
  return 0;
}
