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 = 100100;
struct Hull {
  long long a[MAXN], b[MAXN];
  double x[MAXN];
  int head, tail;
  Hull() : head(1), tail(0) {}
  long long get(long long xQuery) {
    if (head > tail) return 0;
    while (head < tail && x[head + 1] <= xQuery) head++;
    x[head] = xQuery;
    return a[head] * xQuery + b[head];
  }
  void add(long long aNew, long long bNew) {
    double xNew = -1e18;
    while (head <= tail) {
      if (aNew == a[tail]) return;
      xNew = 1.0 * (b[tail] - bNew) / (aNew - a[tail]);
      if (head == tail || xNew >= x[tail]) break;
      tail--;
    }
    a[++tail] = aNew;
    b[tail] = bNew;
    x[tail] = xNew;
  }
};
const int MN = 100111;
int n, m, nFeed;
long long d[MN], p[MN], h[MN], a[MN], s[MN], t[MN];
long long f[111][MN];
int main() {
  while (cin >> n >> m >> nFeed) {
    for (int i = (2), _b = (n); i <= _b; i++) {
      cin >> d[i];
      p[i] = p[i - 1] + d[i];
    }
    for (int i = (1), _b = (m); i <= _b; i++) {
      cin >> h[i] >> t[i];
      a[i] = t[i] - p[h[i]];
    }
    sort(a + 1, a + m + 1);
    for (int i = (1), _b = (m); i <= _b; i++) s[i] = s[i - 1] + a[i];
    for (int i = (1), _b = (nFeed); i <= _b; i++) {
      Hull hull;
      for (int j = (1), _b = (m); j <= _b; j++) {
        f[i][j] = hull.get(a[j]) + a[j] * j - s[j];
        if (i > 1) hull.add(-j, f[i - 1][j] + s[j]);
      }
    }
    cout << f[nFeed][m] << endl;
  }
}
