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;
struct CHT_Min {
  vector<long long> M, C;
  long long pointer;
  void clear() {
    M.clear(), C.clear();
    pointer = 0;
  }
  bool removeCheck(long long id1, long long id2, long long id3) {
    return 1.0 * (C[id3] - C[id1]) * (M[id1] - M[id2]) <
           1.0 * (C[id2] - C[id1]) * (M[id1] - M[id3]);
  }
  void insert_Line(long long m, long long c) {
    if (M.size() > 0 && M.back() == m) return;
    M.push_back(m), C.push_back(c);
    while (M.size() >= 3 &&
           removeCheck(M.size() - 3, M.size() - 2, M.size() - 1)) {
      M.erase(M.end() - 2);
      C.erase(C.end() - 2);
    }
  }
  long long getval(long long id, long long x) { return M[id] * x + C[id]; }
  long long minValPT(long long x) {
    if (pointer >= M.size()) pointer = M.size() - 1;
    while (pointer < M.size() - 1 &&
           M[pointer] * x + C[pointer] >= M[pointer + 1] * x + C[pointer + 1])
      pointer++;
    return M[pointer] * x + C[pointer];
  }
  long long minVal(long long x) {
    long long lo = 0, hi = M.size() - 1;
    long long ans = 1e18;
    while (lo <= hi) {
      long long mid1 = lo + (hi - lo) / 3;
      long long mid2 = hi - (hi - lo) / 3;
      long long v1 = getval(mid1, x);
      long long v2 = getval(mid2, x);
      if (v1 < v2) {
        ans = min(ans, v1);
        hi = mid2 - 1;
      } else {
        ans = min(ans, v2);
        lo = mid1 + 1;
      }
    }
    return ans;
  }
};
long long n, m, p;
long long a[100002], d[100002];
long long dp[102][100002];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(), cout.tie();
  cin >> n >> m >> p;
  for (int i = 2; i <= n; i++) cin >> d[i], d[i] += d[i - 1];
  for (int i = 1; i <= m; i++) {
    long long h, t;
    cin >> h >> t;
    a[i] = t - d[h];
  }
  sort(a + 1, a + m + 1);
  for (int i = 1; i <= m; i++) d[i] = a[i] + d[i - 1];
  for (int i = 1; i <= m; i++) dp[1][i] = a[i] * i - d[i];
  CHT_Min cht;
  for (int i = 2; i <= p; i++) {
    cht.clear();
    for (int j = i; j <= m; j++) {
      cht.insert_Line(-(j - 1), dp[i - 1][j - 1] + d[j - 1]);
      dp[i][j] = cht.minValPT(a[j]) + a[j] * j - d[j];
    }
  }
  cout << dp[p][m] << endl;
  return 0;
}
