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 N = 1e5 + 5, P = 110;
int n, m, p, uk;
const long double inf = 1e15;
struct line {
  long double k, b, l, r;
};
long double itrx(line a, line b) { return (a.b - b.b) / (b.k - a.k); }
vector<line> cht;
void add(long double k, long double b) {
  bool ok = true;
  while (!cht.empty() && ok) {
    if (cht.back().k == k) {
      if (cht.back().b > b)
        cht.pop_back();
      else
        return;
      continue;
    }
    long double x = itrx(cht.back(), {k, b, 0, 0});
    if (x <= cht.back().l)
      cht.pop_back();
    else
      ok = false;
  }
  while (uk >= cht.size()) uk--;
  if (!cht.empty()) {
    long double x = itrx(cht.back(), {k, b});
    cht.back().r = x;
    cht.push_back({k, b, x, inf});
  } else {
    cht.push_back({k, b, -inf, inf});
  }
  uk = max(0, uk);
}
long long d[N], pd[N], k[N], pk[N];
long long dp[P][N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m >> p;
  for (int i = 1; i < n; ++i) cin >> d[i];
  for (int i = 1; i < n; ++i) pd[i] = pd[i - 1] + d[i];
  for (int i = 0; i < m; ++i) {
    int h, t;
    cin >> h >> t;
    k[i + 1] = t - pd[h - 1];
  }
  sort(k + 1, k + m + 1);
  for (int i = 1; i <= m; ++i) pk[i] = pk[i - 1] + k[i];
  for (int i = 1; i <= m; ++i) dp[0][i] = k[i] * i - pk[i];
  for (int i = 1; i < p; ++i) {
    cht.clear();
    cht.push_back({-1, 0, -inf, inf});
    uk = 0;
    for (int j = 1; j <= m; ++j) {
      while (cht[uk].r < k[j]) uk++;
      while (cht[uk].l > k[j]) uk--;
      dp[i][j] = cht[uk].k * k[j] + cht[uk].b + k[j] * (j + 1) - pk[j];
      add(-(j + 1), dp[i - 1][j] + pk[j]);
    }
  }
  cout << dp[p - 1][m];
}
