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>
#pragma GCC optimize("Ofast")
using namespace std;
const long long INF = 1e18;
struct Line {
  long long k, b;
  Line(long long k, long long b) : k(k), b(b) {}
  Line() : k(0), b(0) {}
  pair<long double, long double> intr(Line line) {
    pair<long double, long double> ans;
    ans.first = (long double)(line.b - b) / (k - line.k);
    ans.second = k * ans.first + b;
    return ans;
  }
};
struct CHT {
  vector<long double> stk;
  vector<Line> stl;
  long long get(long long x) {
    long long num =
        lower_bound((stk).begin(), (stk).end(), x) - stk.begin() - 1;
    return stl[num].b + stl[num].k * x;
  }
  bool check(Line l) {
    auto [x, y] = l.intr(stl.back());
    if (x < stk.back()) return true;
    return false;
  }
  void add(Line t) {
    if (!(long long)((stk).size())) {
      stk.push_back(-INF);
      stl.push_back(t);
      return;
    }
    while ((long long)((stk).size()) && check(t))
      stk.pop_back(), stl.pop_back();
    auto [x, y] = t.intr(stl.back());
    stk.push_back(x), stl.push_back(t);
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m, p;
  cin >> n >> m >> p;
  vector<long long> a(m), h(m), t(m), d(n - 1), dp(m, 0), s(n, 0), pref(m, 0);
  for (auto &e : d) cin >> e;
  for (long long i = 0; i < m; i++) cin >> h[i] >> t[i];
  for (long long i = 1; i < n; i++) s[i] = s[i - 1] + d[i - 1];
  for (long long i = 0; i < m; i++) a[i] = t[i] - s[h[i] - 1];
  sort((a).begin(), (a).end());
  pref[0] = a[0];
  for (long long i = 1; i < m; i++) pref[i] = pref[i - 1] + a[i];
  dp[0] = 0;
  for (long long i = 1; i < m; i++) dp[i] = -pref[i] + a[i] * (i + 1);
  for (long long k = 0; k < p - 1; k++) {
    vector<long long> dp2(m, INF);
    dp2[0] = 0;
    CHT kek;
    kek.add({0, pref[0]});
    for (long long j = 1; j < m; j++) {
      dp2[j] = kek.get(a[j]) + a[j] * j - pref[j];
      kek.add({-j, dp[j] + pref[j]});
    }
    swap(dp, dp2);
  }
  cout << dp[m - 1];
}
