Everything is great about Ilya's city, except the roads. The thing is, the only ZooVille road is represented as n holes in a row. We will consider the holes numbered from 1 to n, from left to right.

Ilya is really keep on helping his city. So, he wants to fix at least k holes (perharps he can fix more) on a single ZooVille road. 

The city has m building companies, the i-th company needs ci money units to fix a road segment containing holes with numbers of at least li and at most ri. The companies in ZooVille are very greedy, so, if they fix a segment containing some already fixed holes, they do not decrease the price for fixing the segment. 

Determine the minimum money Ilya will need to fix at least k holes.

Input

The first line contains three integers n, m, k (1 ≤ n ≤ 300, 1 ≤ m ≤ 105, 1 ≤ k ≤ n). The next m lines contain the companies' description. The i-th line contains three integers li, ri, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ci ≤ 109).

Output

Print a single integer — the minimum money Ilya needs to fix at least k holes. 

If it is impossible to fix at least k holes, print -1.

Please, do not use 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

10 4 6
7 9 11
6 9 13
7 7 7
3 5 6


Output

17


Input

10 7 1
3 4 15
8 9 8
5 6 8
9 10 6
1 4 2
1 4 10
8 10 13


Output

2


Input

10 1 9
5 10 14


Output

-1
#include <bits/stdc++.h>
using namespace std;
long long dp[310][310];
long long minv[310][310];
const long long INF = 1000000000000000000LL;
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = (0); i <= (n); i++)
    for (int j = (0); j <= (n); j++) minv[i][j] = INF;
  for (int i = (0); i <= (n); i++)
    for (int j = (0); j <= (n); j++) dp[i][j] = INF;
  for (int i = 0; i < (m); i++) {
    int l, r, c;
    scanf("%d%d%d", &l, &r, &c);
    minv[l][r] = min(minv[l][r], (long long)c);
  }
  for (int l = (n); l >= (1); l--) {
    for (int j = (0); j <= (n - l); j++) {
      minv[j + 1][j + l] = min(minv[j][j + l], minv[j + 1][j + l]);
      minv[j][j + l - 1] = min(minv[j][j + l], minv[j][j + l - 1]);
    }
  }
  dp[0][0] = 0;
  for (int i = (0); i <= (n - 1); i++) {
    for (int j = (0); j <= (n); j++) {
      dp[i + 1][j] = min(dp[i + 1][j], dp[i][j]);
      for (int k = (i + 1); k <= (n); k++) {
        dp[k][j + k - i] = min(dp[k][j + k - i], dp[i][j] + minv[i + 1][k]);
      }
    }
  }
  long long ans = INF;
  for (int j = (k); j <= (n); j++) ans = min(ans, dp[n][j]);
  if (ans == INF) ans = -1;
  printf("%I64d\n", ans);
}
