#include <bits/stdc++.h>
using namespace std;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
T pow(T a, T b, long long m) {
  T ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans % m;
}
vector<pair<long long, long long> > g[400];
long long n, m, k;
long long dp1[400][400];
long long dp2[400][400];
long long solve2(int l, int r) {
  if (l > r) return 0;
  long long ans = 1000000000000000000;
  if (dp2[l][r] != -1) return dp2[l][r];
  for (int i = 0; i < g[l].size(); i++) {
    long long to = g[l][i].first;
    long long cost = g[l][i].second;
    ans = min(ans, cost + solve2(to + 1, r));
  }
  return dp2[l][r] = ans;
}
long long solve1(int id, int dist) {
  if (id == (n + 1) && dist > 0) return 100000000000000000;
  if (dist <= 0) return 0;
  if (dp1[id][dist] != -1) return dp1[id][dist];
  long long ans = 100000000000000000;
  ans = min(ans, solve1(id + 1, dist));
  for (int i = id; i <= n; i++) {
    long long val = solve2(id, i);
    if (val == 1000000000000000000) continue;
    ans = min(ans, val + solve1(i + 1, dist - (i - id + 1)));
  }
  return dp1[id][dist] = ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m >> k;
  long long l, r, d;
  memset(dp1, -1, sizeof dp1);
  memset(dp2, -1, sizeof dp2);
  for (int i = 0; i < m; i++) {
    cin >> l >> r >> d;
    g[l].push_back(make_pair(r, d));
  }
  long long ans = solve1(1, k);
  if (ans == 100000000000000000)
    cout << "-1";
  else
    cout << ans;
}
